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miv72 [106K]
3 years ago
14

You want to push a 71-kg box up a 16° ramp. The coefficient of kinetic friction between the ramp and the box is 0.23. With what

magnitude force parallel to the ramp should you push on the box so that it moves up the ramp at a constant speed?
Physics
1 answer:
ycow [4]3 years ago
8 0

Answer:

<em>345.60N</em>

Explanation:

The magnitude of the force parallel to the ramp that will push the box up the ramp are the frictional force(Ff) and the moving force Fm

Total force = Ff + Fm

Ff = ηR

Ff = ηmgcosθ

Fm = Wsinθ

Fm = mgsinθ

Total force = ηmgcosθ + mgsinθ

Total force = mg(ηcosθ + sinθ)

Given

mass m = 71kg

acceleration due to gravity g = 9.8m/s²

angle of inclination θ = 16°

coefficient of friction η = 0.23

Substitute into the formula;

Total force = 71(9.8)(0.23cos16 + sin16)

Total force = 695.8(0.2211+0.2756)

Total force = 695.8(0.4967)

Total force = 345.60N

<em>Hence the magnitude force parallel to the ramp that you should push on the box so that it moves up the ramp at a constant speed is 345.60N</em>

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A stone was dropped off a cliff and hit the ground with a speed of 88 ft/s. What is the height (in feet) of the cliff
igor_vitrenko [27]

Answer:

the height (in feet) of the cliff is 121 ft

Explanation:

A stone hit the cliff with

speed, v = 88 ft/s

Acceleration, a= 32 ft/s^2

initial speed, u = 0 ft/s

height is h.

To solve this problem we will apply the linear motion kinematic equations, Equation of motion describes change in velocity, depending on the acceleration and the distance traveled

so, writing the formula of Equation of motion:

v^2 - u^2 = 2*a*h

substituting the appropriate values,

(88)^2 - 0 = 2*32* h

h=(88)^2 / 64

h= 121 ft

hence

the height (in feet) of the cliff is 121 ft

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3 0
1 year ago
Hardness and density are physical properties.<br> a. True<br> b. False
antiseptic1488 [7]
a . true hardness and density are physical properties
7 0
2 years ago
Read 2 more answers
See attachment for question
zloy xaker [14]

The Mercury's mass for the given acceleration due to gravity is 0.3152 x 10²⁴ kg.

The ratio of the calculated and accepted value of the Mercury's mass is 0.95.

<h3>What is mass?</h3>

Mass is the amount of matter present in the object.

The mass of the object is always constant, anywhere it is on the Earth or Moon or any other planet.

Given is the acceleration due to gravity of Mercury planet at North pole is g = 3.698 m/s² and the radius of Mercury planet is 2440 km.

The acceleration due to gravity is related with mass as

g = GM/R²

Substitute the values, we have

3.698 = 6.67 x 10⁻¹¹ x M/(2440 x1000)³

M = 2.2016 x 10¹³ /  6.67 x 10⁻¹¹

M = 0.3152 x 10²⁴ kg

Thus, the mercury's mass is  0.3152 x 10²⁴ kg.

(b) Accepted value of Mercury's mass is 3.301 x 10²³ kg

Ratio of the value of mass calculated and accepted is

Mcalc/M accep =  0.3152 x 10²⁴ kg / 3.301 x 10²³ kg

                          = 0.95

Thus, the ratio is 0.95

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8 0
1 year ago
(a) Two ions with masses of 4.39×10^−27 kg move out of the slit of a mass spectrometer and into a region where the magnetic fiel
sammy [17]

Answer:

Part a)

R_1 = 0.072 m

Part b)

R_2 = 0.036 m

Part c)

d = 0.072 m

Explanation:

Part a)

As we know that the radius of the charge particle in constant magnetic field is given by

R = \frac{mv}{qB}

now for single ionized we have

R_1 = \frac{(4.39\times 10^{-27})(7.92 \times 10^5)}{(1.6 \times 10^{-19})(0.301)}

R_1 = 0.072 m

Part b)

Similarly for doubly ionized ion we will have the same equation

R = \frac{mv}{qB}

R_2 = \frac{(4.39\times 10^{-27})(7.92 \times 10^5)}{(3.2 \times 10^{-19})(0.301)}

R_2 = 0.036 m

Part c)

The distance between the two particles are half of the loop will be given as

d = 2(R_1 - R_2)

d = 2(0.072 - 0.036)

d = 0.072 m

6 0
3 years ago
A copper wire of length 16 cm is in a magnetic field of 0.19 T. If it has a mass of 9.0 g, what is the minimum current through t
Sergeeva-Olga [200]

Answer:

2.9 A

Explanation:

L = 16 cm = 0.16 m

B = 0.19 T

m = 9 g = 0.009 kg

Let the minimum current be i.

Magnetic force is balanced by the gravitational force

B x i x L = m x g

0.19 x i x 0.16 = 0.009 x 9.8

i = 2.9 A

7 0
3 years ago
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