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Yanka [14]
3 years ago
11

A truck travels up a hill with a 5.7° incline. The truck has a constant speed of 22 m/s. What is the horizontal component of the

truck’s velocity? Answer in units of m/s.

Physics
1 answer:
tester [92]3 years ago
8 0

Answer:

The horizontal component of the velocity is 21.9 m/s.

Explanation:

Please see the attached figure for a better understanding of the problem.

Notice that the vector v and its x and y-components (vx and vy) form a right triangle. Then, we can use trigonometry to find the magnitude of vx, the horizontal component of the velocity.

To find vx, let´s use the following trigonometric rule of right triangles:

cos α = adjacent / hypotenuse

cos 5.7° = vx / 22 m/s

22 m/s · cos 5.7° = vx

vx = 21.9 m/s

The horizontal component of the velocity is 21.9 m/s.

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irina1246 [14]

Answer:

A.Gravity acts to pull the object down

D.The object’s inertia carries it forward.

E.The path of the object is curved.

Explanation:

The motion of a projectile consists of two independent motions:

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- A vertical motion with constant acceleration of g = 9.8 m/s^2 downward (acceleration due to gravity), due to the presence of the force of gravity, so the vertical velocity changes (increases in the downward direction)

As a result, the combined motion of the projectile has a curved trajectory (parabolic, more specifically). So the following options are correct:

A.Gravity acts to pull the object down --> gravity acts along the vertical direction

D.The object’s inertia carries it forward. --> there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal motion continues with constant speed

E.The path of the object is curved

3 0
4 years ago
Why are the Soviets ahead in the race to capture the V-2 rocket and Werner von Braun?
Dahasolnce [82]

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With this information, we can say that because it was a base discovered by US troops, none of the alternatives is correct, as it was not the Soviets who discovered it, and that the base was also located in the central part of Germany.

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5 0
3 years ago
A 0.473 kg ice puck, moving east with a speed of 2.76 m/s, has a head-on collision with a 0.819 kg puck initially at rest. Assum
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Answer:

The final speed of puck 1 is 0.739 m/s towards west  and puck 2 is 2.02 m/s towards east .

Explanation:

Let us consider east as positive direction and west as negative direction .

Given

mass of puck 1 , m_1= 0.473 kg

mass of puck 2 , m_2= 0.819 kg

initial speed of puck 1 , u_1=2.76\frac{m}{s}

initial speed of puck 2 , u_2=0.00\frac{m}{s}

Final speed of puck 1 and puck 2 be v_1\, and\, v_2  respectively

Apply conservation of linear momentum

m_1u_1+m_2u_2=m_1v_1+m_2v_2

=>0.473\times 2.76+0.0=0.473\times v_1+0.819\times v_2

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Since collision is perfectly elastic , coefficient restitution e=1

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From equation (A) and (B)

v_1=-0.739\frac{m}{s}

and v_2=2.02\frac{m}{s}

Thus the final speed of puck 1 is 0.739 m/s towards west  and puck 2 is 2.02 m/s towards east .

       

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Subtract n from each side
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