A motor vehicle has a mass of 1.8 tonnes and its wheelbase is 3 m. The centre of gravity of the vehicle is situated in the central plane 0.9 m above the ground and 1.7 m behind the front axle. When moving on the level at 90 km/h the brakes applied and it comes to a rest in a distance of 50 m.
Calculate the normal reactions at the front and rear wheels during the braking period and the least coefficient of friction required between the tyres and the road. (Assume g = 10 m/s2)​

Answer:

A) receding from the earth

B) [tex]3.078x10^6m/s[/tex]

Explanation:

The wavelength went from 434.1nm to 438.6nm, there was an increase in wavelength (also knowecn as redshift due to the doppler efft), this increase is due to the fact that the source that emits the radiation (the distant galaxy) is moving away and therefore the light waves it emits are "stretched", causing us to see a wavelength greater than the original.

to calculate the relative speed we use the following formula:

[tex]v_{rel}=c(1-\frac{\lambda_{1}}{\lambda_{2}} )[/tex]

where [tex]c[/tex] is the speed of light: [tex]c=3x10^8m/s[/tex]

[tex]\lambda_{1}[/tex] is the wavelength emited by the source, and

[tex]\lambda_{2}[/tex] is the wavelength measured on earth.

we substitute all the values and do the calculations:

[tex]v_{rel}=(3x10^8m/s)(1-\frac{434.1nm}{438.6nm} )\\\\v_{rel}=(3x10^8m/s)(1-0.98974)\\\\v_{rel}=(3x10^8m/s)(0.01026)\\\\v_{rel}=3.078x10^6m/s[/tex]

the relative speed is: [tex]3.078x10^6m/s[/tex]

Answer:

???????

Explanation:

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Answer:

The Answer is red is the lowest and violet is the highest frequency

Explanation:

I think that means A, because the red isn't in the question. But I'm sure red is the lowest frequency and violet is the highest in the visible light spectrum

The visible spectrum as it appears to the human eye is that A. the lowest frequency appears red, and the highest frequency appears violet.

Humans can only view a portion of the electromagnetic spectrum and this portion is known as visible light.

The colors in this visible light have different frequencies which include:

Notice how red is the lowest frequency and violet is the highest so we can conclusively say that the lowest frequency appears red, and the highest frequency appears violet.

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Answer:

176.38 rpm

Explanation:

mass percentage of arms and legs = 13%

mass percentage of legs and trunk = 80%

mass percentage of head = 7%

Total mass of the skater = 74.0 kg

length of arms = 70 cm = 0.7 m

height of skater = 1.8 m

diameter of trunk = 35 cm = 0.35 m

Initial angular momentum = 68 rpm

We assume:

We split her body into two systems, the spinning hands as spinning rods

1. Each rod has moment of inertia = [tex]\frac{1}{3} mL^{2}[/tex]

mass m of the arms is 13% of 74 kg = 0.13 x 74 = 9.62 kg

mass of each side will be assumed to be 9.62/2 = 4.81 kg

L = length of each arm

therefore,

I =  [tex]\frac{1}{3}[/tex] x 4.81 x [tex]0.7^{2}[/tex] = 0.79 kg-m   for each arm

2. Her body as a cylinder has moment of inertia =  [tex]\frac{1}{2} mr^{2}[/tex]

r = radius of her body = diameter/2 = 0.35/2 = 0.175 m

mass of body trunk = (80% + 7%) of 74 kg = 0.87 x 74 = 64.38 kg

I = [tex]\frac{1}{2}[/tex] x 64.38 x [tex]0.175^{2}[/tex] = 0.99 kg-m

We consider each case

case 1: Body spinning with arm outstretched

Total moment of inertia = sum of moments of inertia of both arms and moment of inertia of body trunk

I = (0.79 x 2) +  0.99 = 2.57 kg-m

angular momentum = Iω

where ω = angular speed = 68.0 rpm = [tex]\frac{2\pi }{60}[/tex] x 68 = 7.12 rad/s

angular momentum = 2.57 x 7.12 = 18.29 kg-rad/m-s

case 2: Arms pulled down parallel to trunk

The momentum of inertia will be due to her body trunk alone which is 0.91 kg-m

angular momentum =  Iω

=  0.99 x ω = 0.91ω

according to conservation of angular momentum, both angular momentum must be equal, therefore,

18.29 = 0.99ω

ω = 18.29/0.99 = 18.47 rad/s

18.47 ÷ [tex]\frac{2\pi }{60}[/tex]  = 176.38 rpm

Explanation:

A substance with a short , medium, free path has improved electron flow resistance and a higher electrical resistance . Heat applications impose more molecular chaos on all materials and shorten the track further, increasing resistance of most materials. So, just refresh the material to expand the course. In certain materials, when cooled to the minimum temperature, the conductivity is substantially increased.

Answer:

P = I²r

Explanation:

ε= IR + Ir

where r is the internal resistance

Answer:

(a) Area(small piston)/Area(large piston) = 0.037

(b) h = 1336.36 cm = 13.36 m

Explanation:

(a)

The stress on the smaller piston is equally transmitted to the larger piston, in a hydraulic lift. Therefore,

Stress (small piston) = Stress (large piston)

Force (small piston)/Area (small piston) = Force (Large Piston)/Area (Large Piston)

Area(small piston)/Area(large piston) = Force (small piston)/Force(Large piston)

Area(small piston)/Area(large piston) = 550 N/(1500 kg)(9.8 m/s²)

Area(small piston)/Area(large piston) = 0.037

(b)

The work is also transmitted equally to the large piston. So,

Work(small piston) = Work(Large Piston)

Force(small piston).Displacement(small piston) = Force(large piston).Displacement(small piston)

(550 N)(h) = (1500 kg)(9.8 m/s²)(50 cm)

h = 735000 N.cm/550 N

h = 1336.36 cm = 13.36 m

(a) The ratio of area smaller piston to area of larger piston is 0.037.

(b) The distance the smaller piston will be pushed down to lift the car at the given height is 13.36 m.

The given parameters;

let the area of the small piston = A₁

let the area of the large piston = A₂

Apply constant pressure principle as shown below;

[tex]P = \frac{F_1}{A_1} = \frac{F_2}{A_2} \\\\\frac{A_1}{A_2} = \frac{A_{small}}{A_{large}} = \frac{F_1}{F_2} = \frac{550}{mg} \\\\ \frac{A_{small}}{A_{large}} = \frac{550}{1500 \times 9.8} \\\\ \frac{A_{small}}{A_{large}} = 0.037[/tex]

The distance the effort will be applied is calculated as follows;

[tex]550 d = mgh\\\\550d = (1500 \times 9.8 \times 0.5)\\\\550 d = 7350\\\\d = \frac{7350}{550} \\\\d = 13.36 \ m[/tex]

Thus, the distance the smaller piston will be pushed down to lift the car at the given height is 13.36 m.

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Answer:

Explanation:

The parametric equations of the cyclist are:

[tex]x(t)=100tcos(A)\\\\y(t)=-16t^2+100tsin(A)+10[/tex]   (1)

A) The horizontal velocity is the derivative of x(t), in time:

[tex]x'(t)=100cos(A)[/tex]

B) For A=20° the horizontal velocity is:

[tex]v_x=x'(t)=100cos(20\°)=93.9692ft/s[/tex]

For A=45°:

[tex]v_x=100cos(45\°)=70.7106ft/s[/tex]

C) To find the time in which the vertical velocity is zero you first obtain the derivative of, in time:

[tex]v_y=y'(t)=-32t+100sin(A)+10[/tex]

Next, you equal the vertical velocity to zero and solve for time t:

[tex]-32t+100sin(A)+10=0\\\\t=\frac{100sin(A)+10}{32}[/tex]

D) The maximum height is reached when the derivative of y (height) is zero. You use the previous value of t in the equation (1), equals y to 35. Next, you  solve for t:

[tex]y=35\\\\-16(\frac{100sin(A)+10}{32})^2+100(\frac{10sin(A)+10}{32})sinA+10=35\\\\-\frac{16}{1024}(10000sin^2A+2000sinA+100)+31.25sin^2A+31.25sinA+10=35\\\\-156.25sin^2A-31.25sinA-1.5625+31.25sin^2A+31.25sinA+10=35\\\\-125sin^2A-26.5625=0[/tex]

Answer:

Step 1

Work done = -9134.4 J

ΔQ = -9134.4 J

Step 2

ΔQ =  -3570.32 J = ΔU

W = 0

Step 3

The pdV work done = 3570.32 J

The Vdp work done = 11053.37 J

Heat transferred, ΔE = 0.

Explanation:

For diatomic gases γ = 1.4

Step 1

Where:

v₂ = 3.00·v₁

On isothermal expansion of an ideal gas by Boyle's law, we have;

p₁·v₁ = p₂·v₂ which gives;

p₁·v₁ = p₂×3·v₁

Dividing both sides by v₁, we have;

p₁= 3·p₂

[tex]p_2 = \dfrac{p_1}{3}[/tex]

Hence, the pressure is reduced by a factor of 3

Work done =

[tex]n\cdot R\cdot T\cdot ln\dfrac{v_{f}}{v_{i}}[/tex]

Where:

n = 1 mole  

R = 8.3145 J/(mole·K)

T = 1000 K we have

[tex]1 \times 8.3145 \times 1000 \times ln\left (\dfrac{1}{3} \right ) = -9134.4 J[/tex]

Step 2

The gas undergoes a constant volume decrease in pressure given by Charles law as follows;

[tex]\dfrac{p_2}{p3} = \dfrac{T_1}{T_3}[/tex]

Whereby p₂ > p₃, T₁ will be larger than T₃

W = 0 for constant volume process

ΔQ = m×cv×ΔT = 1 × 3.97 × -900 = -3570.32 J = ΔU

Step 3

For adiabatic compression, we have;

[tex]\dfrac{p_3}{p_1} = \left (\dfrac{V_1}{V_3} \right )^{\gamma } = \left (\dfrac{T_3}{T_1} \right )^{\frac{\gamma }{\gamma -1}}[/tex]

Where:

T₁ = 1000 K

T₃ = 100 K

We have;

[tex]\left (\dfrac{V_1}{3\cdot V_1} \right )^{\gamma } = \left (\dfrac{100}{1000} \right )^{\dfrac{\gamma}{\gamma -1}}[/tex]

[tex]\left (\dfrac{1}{3} \right ) = \left (\dfrac{1}{10} \right )^{\dfrac{1}{\gamma -1}}[/tex]

[tex]log\left (\dfrac{1}{3} \right ) = {\dfrac{1}{\gamma -1}} \times log \left (\dfrac{1}{10} \right )^[/tex]

[tex]\gamma -1 =\dfrac{log \left (\dfrac{1}{10} \right )}{ log\left (\dfrac{1}{3} \right ) } {[/tex]

∴ γ-1 = 2.096

γ = 3.096

The pdV work done =

[tex]m \times c_v \times (T_1 - T_3)[/tex]

m×R/(γ - 1)×(T₁ - T₃) =

3.97×(1000 - 100) = 3570.32 J

The Vdp work done =

[tex]m \times c_p \times (T_1 - T_3)[/tex]

[tex]c_p = k \times c_v = 3.096 \times 3.97 = 12.3 \, J/(mol\cdot K)[/tex]

12.3×(1000 - 100) = 11053.37 J

Heat transferred, ΔE = 0.

⛔⛔⛔⛔⛔⛔⛔⛔⛔⛔

Answer:

[tex]c_B=0.0768\frac{kcal}{kg\cdot\°C}[/tex]

Explanation:

In order to calculate the specific heat of the material B, you use the following formula for the change in the temperature of a substance, where an amount of heat Q is given to the substance:

[tex]Q=mc(T_2-T_1)[/tex]

Q: amount oh heat

m: mass of the substance

T2: final temperature

T1: initial temperature

c: specific heat of the substance.

If QA and QB are the heat of material A and B, you have:

[tex]Q_A=m_Ac_A(T_{2A}-T_{1A})\\\\Q_B=m_Bc_B(T_{2B}-T_{1B})[/tex]

both materials have the same mass, mA = mB

cA: specific heat of A = 0.2007 kcal/(kg.°C)

cB: specific heat of B = ?

T2A: final temperature of A = 86.3°C

T1A: initial temperature of A = 22.0°C

T2B: final temperature of B = 190.0°C

T1B: initial temperature of B = 22.0°C

In this case you have that both material A and B receive the same amount of heat Q. Then, you equal QA with QB and solve for cB:

[tex]m_Ac_A(T_{2A}-T_{1A})=m_Bc_B(T_{2B}-T_{1B})\\\\c_B=\frac{c_A(T_{2A}-T_{1A})}{(T_{2B}-T_{1B})}\\\\c_B=\frac{(0.2007kcal/(kg.\°C))(86.3\°C-22.0\°C)}{190.0\°C-22.0\°C}\\\\c_B=0.0768\frac{kcal}{kg\cdot\°C}[/tex]

hence, the specific heat of the second rod B is 0.0768kcal/(kg°C)

Answer:

Work Done = 67.5 J

Explanation:

First we find the value of spring constant (k) using Hooke's Law. Hooke's is formulated as:

F = kx

where,

F = Force Applied = 450 N

k = Spring Constant = ?

x = Stretched Length = 30 cm = 0.3 m

Therefore,

450 N = k(0.3 m)

k = 450 N/0.3 m

k = 1500 N/m

Now, the formula for the work done in stretching the spring is given as:

W = (1/2)kx²

Where,

W = Work done = ?

k = 1500 N/m

x = 70 cm - 40 cm = 0.3 m

Therefore,

W = (1/2)(1500 N/m)(0.3 m)²

W = 67.5 J

Answer:

The weight of the pole can be assumed to act at the pole's midpoint, which is 2m from the fence / pivot point, giving us a moment of 490Nm (245N x 2m).  We have to counteract this moment by holding the end of the pole.  So, we have a lever arm of 3.65m (4.0m - 0.35m), so we would need to exert a force of 134.4N (490N / 3.65m) at a point 35cm from the end of the pole.

Explanation:

happy to help:)

Answer:

Explanation:

Let mass of ice cube taken out be m kg .

ice will gain heat to raise its temperature from - 5.5° to 0° and then from 0° to 17° .

Total heat gained = m x 2.1 x 5.5 + m x 333 + m x 4.186 x 17

= (11.55 + 333 + 71.162 )m

= 415.712 m kJ

Heat lost by aluminium calorimeter

= .075 x .9 x 3  

= .2025 kJ

Heat lost by water

= .3 x 4.186 x 3

= 3.7674

Total heat lost

= 3.9699 kJ

Heat lost = heat gained

415.712 m = 3.9699

m = .0095 kg

9.5 gm .

Answer:

0.00954g or 9.5x[tex]10^{-3}[/tex] kg

Explanation:

The only conversion that is needed is changing the heat of fusion of water from 333 kJ/kg to 333000 J/kg.

This is the condensed version of the equation needed for this problem: mcΔT + mL + mcΔT = mcΔT + mcΔT

This is the expanded version of the equation needed for this problem:

[tex]m_{ice}[/tex]([tex]c_{ice}[/tex])(temperature of ice from -5.5°C to 0°C) + [tex]m_{ice}[/tex](L) + [tex]m_{ice}[/tex]([tex]c_{water}[/tex])(temperature of water from 0°C to 17°C) = [tex]m_{water}[/tex]([tex]c_{water}[/tex])(ΔT) + [tex]m_{aluminum}[/tex]([tex]c_{aluminum}[/tex])(ΔT)

Use the equation to solve for the mass of ice:

m(2100)(5.5) + m(333000) + m(4186)(17) = 0.3(4186)(20-17) + 0.075(900)(20-17)

m [(2100x5.5) + 333000 + (4186x17)] = 3767.4 + 202.5

m(415712) = 3969.9

m = 0.00954g or 9.5x[tex]10^{-3}[/tex] kg

Answer:

The acceleration of [tex]M_2[/tex] is  [tex]a = 0.7156 m/s^2[/tex]

Explanation:

From the question we are told that

    The mass of first block is  [tex]M_1 = 2.25 \ kg[/tex]

    The angle of inclination of first block is  [tex]\theta _1 = 43.5^o[/tex]

    The coefficient of kinetic friction of the first block is  [tex]\mu_1 = 0.205[/tex]

      The mass of the second block is  [tex]M_2 = 5.45 \ kg[/tex]

     The angle of inclination of the second block is  [tex]\theta _2 = 32.5^o[/tex]

      The coefficient of kinetic friction of the second block is [tex]\mu _2 = 0.105[/tex]

The acceleration of [tex]M_1 \ and\ M_2[/tex] are same

The force acting on the mass [tex]M_1[/tex] is mathematically represented as

     [tex]F_1 = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1[/tex]

=> [tex]M_1 a = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1[/tex]

Where T is the tension on the rope

The force acting on the mass [tex]M_2[/tex] is mathematically represented as    

  [tex]F_2 = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2[/tex]

   [tex]M_2 a = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2[/tex]

At equilibrium

  [tex]F_1 = F_2[/tex]

So

 [tex]T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2[/tex]

making a the subject of the formula

    [tex]a = \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}[/tex]

substituting values [tex]a = \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}[/tex]

    => [tex]a = 0.7156 m/s^2[/tex]

The acceleration of the second block to the right is 2.21 m/s².

The normal force on block1 is calculated as follows;

[tex]F_n_1 = m_1g cos(\theta_1)[/tex]

The parallel force on block 1 is calculated as;

[tex]F_x_1 = m_1gsin(\theta)[/tex]

The frictional force on block 1 is calculated as;

[tex]F_k_1 = \mu_k F_n = \mu_k m_1gcos\theta_1[/tex]

The net force on block 1 is calculated as;

[tex]\Sigma F_x_1 = m_1gsin(\theta_1) - \mu_k_1m_1gcos(\theta_1)[/tex]

The normal force on block 2 is calculated as follows;

[tex]F_n_2 = m_2gcos\theta _2[/tex]

The frictional force on block 2 is calculated as;

[tex]F_k_2 = \mu k_2 m_2g cos\theta _2[/tex]

The net force on block 2 is calculated as follows;

[tex]\Sigma F_x_2 = m_2a_2\\\\ m_2g_2 sin(\theta _2) - F_k_2 - \Sigma F_x_1 = m_2a_2 \\\\m_2gsin(\theta) - F_k_2 - (m_1gsin(\theta) - \mu_k _1 m_1g cos(\theta)) = m_2a_2\\\\m_2gsin(\theta) -\mu_k_2 m_2gcos(\theta) + \mu_k _1 m_1g cos(\theta) - m_1gsin(\theta) = m_2a_2\\\\5.45( 9.8) sin(32.5) -(0.105)(5.45)(9.8)cos(32.5) + \\\\0.205( 2.25) ( 9.8)cos(43.5) - 2.25( 9.8) sin (43.5) = 5.45a_2\\\\ 12.07 = 5.45a_2\\\\a_2= \frac{12.07}{5.45} \\\\a_2 = 2.21 \ m/s^2[/tex]

Thus, the acceleration of the second block to the right is 2.21 m/s².

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Answer:

Explanation:

We shall apply Stefan's formula

E = AσT⁴

When T = 300

I₁ = Aσ x 300⁴

When T = 400K

I₂ = Aσ x 400⁴

I₂ / I₁ = 400⁴ / 300⁴

= 256 / 81

= 3.16

I₂ = 3.16 I₁ .

Answer:

Option D. is correct

Explanation:

Daniel Levinson, a psychologist developed an adult development theory. This theory was referred to as the Seasons of Life theory. This theory describes the stage in which a person leaves adolescence and start making choices about adult life.

Daniel Levinson found that men and women go through the same stages of development, differ in terms of their social roles and identities, and deal with the development tasks in each stage differently.

Option D. is correct

Answer:

Daniel Levinson found that men and women

d.  all of the above

Explanation:

cause i got a 100 on edg;)

Answer:

F = 1.2×10⁻³ N

Explanation:

From the question,

Applying newton's second law of motion,

F = m(v-u)/t................... Equation 1

Given: F = magnitude of the average force exerted on the ball, m = mass of the ball, v = final velocity, u = initial velocity, t = time of  contact.

Note: let downward be negative and upward be positive.

Given: m = 48 g = 48/1000 = 0.048 kg, v = 17 m/s, u = -28 m/s (downward),

t = 1800 s

Substitute into equation 1

F = 0.048(17-[28])/1800

F = 1.2×10⁻³ N

Answer:

1) We can estimate the age of the earth

2) we can calculate the speed of anything

3) we can also calculate gravity, e.t.c.

Explanation:

I could give you more just ask

Answer:

Δt = 0.315s

Explanation:

To calculate the time difference, in which both fans hear the batterstrike, you first calculate the time which takes the sound to travel the distances to both fans:

[tex]t_1=\frac{d_1}{v_s}[/tex]

[tex]t_2=\frac{d_2}{v_s}[/tex]

d1: distance to the first fan = 18.3 m

d2: distance to the second fan = 127 m

vs: speed of sound = 345 m/s

You replace the values of the parameters to calculate t1 and t2:

[tex]t_1=\frac{18.3m}{345m/s}=0.053s\\\\t_2=\frac{127m}{345m/s}=0.368s[/tex]

The difference in time will be:

[tex]\Delta t =t_2-t_2=0.368s-0.053s=0.315s[/tex]

Hence, the time difference between hearing the sound at the location s of both fans is 0.315s

Answer:

799.54 ft

Explanation:

Linear thermal expansion is:

ΔL = α L₀ ΔT

where ΔL is the change in length,

α is the linear thermal expansion coefficient,

L₀ is the original length,

and ΔT is the change in temperature.

Given:

α = 1.2×10⁻⁵ / °C

L₀ = 800 ft

ΔT = -17°C − 31°C = -48°C

Find: ΔL

ΔL = (1.2×10⁻⁵ / °C) (800 ft) (-48°C)

ΔL = -0.4608

Rounded to two significant figures, the change in length is -0.46 ft.

Therefore, the final length is approximately 800 ft − 0.46 ft = 799.54 ft.

Answer:1.89 m

Explanation:

Given

Block travels [tex]0.63\ m[/tex] in first second

It is released from rest i.e. initial speed is zero (u=0)

using

[tex]s=ut+\frac{1}{2}at^2[/tex]

where a=acceleration

here acceleration is the component of gravity on incline plane (say [tex]\theta [/tex])

so

[tex]s_1=\frac{1}{2}\times g\sin \theta (1)^2[/tex]

[tex]0.633\times 2=9.8\sin \theta \times 1^2[/tex]

[tex]\sin\theta =0.1291[/tex]

[tex]\theta =7.41^{\circ}[/tex]

So distance traveled in [tex]2\ sec[/tex]

[tex]s=\frac{1}{2}\times g\sin \theta (2)^2[/tex]

[tex]s=0.5\times 9.8\times \sin (7.41)\times 4[/tex]

[tex]s=2.52\ m[/tex]

So distance traveled in [tex]2^{nd}\ sec[/tex] is

[tex]s-s_1=2.52-0.633=1.89\ m[/tex]

Answer:

0.4778 m/s

Explanation:

To solve this question, we will make use of law of conservation of momentum.

We are given that the rock's velocity is 12 m/s at 35°. Thus, the horizontal component of this velocity is;

V_x = (12 m/s)(cos(35°)) = 9.83 m/s.

Thus, the horizontal component of the rock's momentum is;

(3.5 kg)(9.83 m/s) = 34.405 kg·m/s.

Since the person is not pushed up off the ice or down into it, his momentum will have no vertical component and so his momentum will have the same magnitude as the horizontal component of the rock's momentum.

Thus, to get the person's speed, we know that; momentum = mass x velocity

Mass of person = 72 kg and we have momentum as 34.405 kg·m/s

Thus;

34.405 = 72 x velocity

Velocity = 34.405/72

Velocity = 0.4778 m/s

Answer: acceleration = 3.27m/s^2

Explanation:

Given that the

Mass M = 44kg

Angle Ø = 15 degree

Coefficient of friction ų = 0.7

Force F = 222N

F - Fr = ma ...... (1)

Where Fr = frictional force

Fr = ųN

N = normal reaction = mg

Fr = ųmgsinØ

Fr = 0.7 × 44 × 9.81 × sin 15

Fr = 78.2N

Substitutes Fr, F and M into equation one.

222 - 78.2 = 44a

143.79 = 44a

Make a the subject of formula

a = 143.79/44

Acceleration a = 3.27 m/s^2

Answer:

right is the correct answer to the given question .

Explanation:

In this question figure is missing

The main objective right-hand rule to decide the position of the magnetic force on the positive force acting, either the position of the thumb of a right hand with in position of v, the fingers throughout the position of B1, and a right angles throughout the position of F1 to the hand positions.

So    [tex]F1 \ =\ q\ v \ B1\ Sin\alpha[/tex]

Answer:

Right

Explanation:

Just did this on edgen, in the diagram, the fingers are pointing to the right, which indicates that the magnetic field is acting to the right.

Answer:

1. v =  22.2 m/s

2. t = 2.25 seconds

3. h = 27.05 m

4. t = 1.16 seconds

Explanation:

The questions involve motion under the influence of gravity

1. Using the formula v² = u² + 2gh

where u = 1.6 m/s; g= 9.81 m/s²; h = 25 m; v = ?

v² = (1.6)² + 2 * 9.81 * 25

√v² = √493.06

v =  22.2 m/s

2. Using h = ut + 1/2 gt²

where h = 25 m; u = 0 (since velocity on reaching the ground is zero); g = 9.81 m/s²; t = ?

therefore, h = 1/2 gt²

making t subject of the formula, t = √ (2*h /g)

t = √ (2 * 25 / 9.81)

t = 2.25 seconds

3. Time of travel for the blue ball, t = 2 - 0.4 = 1.6s

using h = ut - gt²

u = 24 m/s; t = 1.6 s; g = 9.81 m/s²

note: since the ball is travelling against gravity, g is negative

h = 24 * 1.6 - 11/2 * 9.81 * 1.6²

h = 38.4 - 12.55 = 25.85 m

since height above the ground is 1.2 m,

total height h = 25.85 m + 1.2 m

h = 27.05 m

4. Let the time of travel of the red ball be t seconds.

So the time of travel of the blue ball = (t - 0.4) seconds.

Both the balls are at the same height :

25 - s = 1.2 + h  where s & h are the displacements of the red & the blue ball respectively.

25 - (ut + 1/2 gt2) = 1.2 + (ut - 1/2 gt2)

25 - (1.6 t + 0.5 * 9.8 t²) = 1.2 + (24(t-0.4) - 0.5*9.8*(t-0.4)²)

solving the equation above for the time after which both the balls are at the same height.

25 - 1.6t - 4.9t² = 1.2 + 24t - 9.6 - 4.9t² + 3.92t - 0.784

collecting like terms

(25 - 1.2 + 9.6 + 0.784) = (24 + 3.92 + 1.6) * t  

t = 34.184 / 33.44

t = 1.16 seconds

Answer:

0.795 kg

Explanation:

Assuming the complete question:

A person with compromised pinch strength in their fingers can only exert a normal force of 6.0 N to either side of a pinch-held  object. What is the mass of the heaviest book this person can hold onto vertically before it slips out of his or her fingers? The  coefficient of static friction of the surface between the fingers and the book cover is 0.60

SOLUTION:

The maximum weight of the book will equal the maximal friction force that can be produced:

m g = 2 f [tex]F{normal}[/tex]

Note that there are two sides of the book, so the friction force equals 2 times the friction force on one side (hence the factor 2).

So the maximum mass of the book is

m = 2 f [tex]F_n[/tex]/ g

m = 2[tex]\times[/tex]0.65[tex]\times[/tex] 6.0N / (9.81N/kg)  

m = 0.795 kg

Answer:

The image is virtual

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