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Ber [7]
3 years ago
6

An open container of oil rests on the flatbed of a truck that is traveling along a horizontal road at 55 mi/hr. As the truck slo

ws uniformly to a complete stop in 5 s, what will be the slope of the oil surface during the period of constant deceleration

Physics
1 answer:
rjkz [21]3 years ago
3 0

Answer:

Slope = 0.50.

Explanation:

Below is an attachment containing the solution to the question.

You might be interested in
David runs 100 m in a time of 9.80 s. Calculate his average speed. State the unit.
fiasKO [112]

Answer:

His average speed is 980 units

3 0
3 years ago
A 980 kg roller coaster cart is traveling along a track at 17 m/s before it rolls down a 30 m tall hill (Point A). What will be
MrRissso [65]

The kinetic energy halfway the hill is 2.86\cdot 10^5 J

Explanation:

If there are no friction forces acting on the cart, we can apply the law of conservation of energy: the mechanical energy of the cart (which is the sum of potential energy + kinetic energy) must be conserved. So we can write:

U_A +K_A = U_B + K_B

where

U_A=mgh_A is the initial potential energy, at point A, with

m = 980 kg (mass of the cart)

g=9.8 m/s^2 (acceleration of gravity)

h_A = 30 m (height at point A)

K_A=\frac{1}{2}mv_A^2 is the initial kinetic energy, at point A , with

v_A=17 m/s (velocity at point A)

U_B=mgh_B is the final potential energy, at point B, where

h_B = 15 m (height at point B)

K_B=\frac{1}{2}mv_B^2 is the final kinetic energy, at point B, where

v_B is the velocity at point B

Here we are interested in finding K_B, so by re-arranging the equation and substituting we find:

K_B = U_A+K_B-U_B = mg(h_A-h_B)+\frac{1}{2}mv_A^2=(980)(9.8)(30-15)+\frac{1}{2}(980)(17)^2=2.86\cdot 10^5 J

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

8 0
3 years ago
How is a quarter able to role on its edge?
Olegator [25]
Idk its really weird because the ridges are making the quarter stand and therefore able to roll
5 0
3 years ago
a 4m long straight wir that carries acurrent of 0.5A is placed perpendicular to a uniform magnetic field. if the size of magneti
PolarNik [594]

Answer:

B=0.2T

Explanation:

given required solution

l=4m B=? <em>F</em><em>=</em><em>BIL</em>

i=0.5A B=F/IL

F=0.4N B=0.4N/0.5A*4m

B=0.4/2=0.2T

5 0
3 years ago
Help please I’ll mark as brainliest
kirza4 [7]

Answer:

yes

Explanation:

6 0
2 years ago
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