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Advocard [28]
3 years ago
6

A roller coaster's velocity at the top of the hill is 10 m/s. Two seconds later it reaches the bottom of the hill with a velocit

y of 26m/s. What is the acceleration of the coaster?
Physics
1 answer:
svet-max [94.6K]3 years ago
5 0

your answer is 8 m / sec²

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What is the magnitude of the velocity of a 25 kg mass that is moving with a momentum of 100 kg*m/s?
Gekata [30.6K]

Answer:

v= 4 m/s

Explanation:

Momenutm is, by definition, the product of mass and velocity.

p = mv

Let's replace what we know and solve for whatever's left

100 kg\cdot m/s = 25kg \cdot v \rightarrow v= 4 m/s

7 0
3 years ago
The two cars collide at right angles in the intersection of two icy roads. Car A has a mass of 1965 kg and car B has a mass of 1
Sunny_sXe [5.5K]

Answer:

U2 = 47.38m/s = initial velocity of B before impact

Explanation:

An example of the diagram is shown in the attached file because of missing angle of direction in the question

Mass A, B are mass of cars

A = 1965

B =1245

U1 = initial velocity of A = 52km/hr

U2 = initial velocity of B

V = common final velocity of two cars

BU2 = (A + B)*V sin ¤ ...eq1 y plane

AU1 = (A + B) *V cos ¤ ....equ 2plane

From equ 2

V = AU1/(A + B)*cos ¤

Substitute V into equation 1

We have

U2 = (AU1/B)tan ¤ where ¤ = angle of direction which is taken to be 30°

Substitute all parameters to get

U2 = (1965/1245)*52 * tan 30°

U2 = 47.38m/s

8 0
4 years ago
How would you define compound
aliina [53]

Answer:

as a way to put two things together and yeeah

Explanation:

i know because this is what my mom told me

5 0
3 years ago
I MARK BRAINLIEST, PLEASE ANSWER ASAP!!! PHYSICS 20 POINTS!!
irina [24]
The first one is C. Hope this helps!! If you need anymore help just message me and I will try to get back to you quickly and help in any way i can!
5 0
3 years ago
Read 2 more answers
P-weight blocks D and E are connected by the rope which passes through pulley B and are supported by the isorectangular prism ar
creativ13 [48]

Answer:

21.8°

Explanation:

Let's call θ the angle between BC and the horizontal.

Draw a free body diagram for each block.

There are 4 forces acting on block D:

Weight force P pulling down,

Normal force N₁ pushing perpendicular to AB,

Friction force N₁μ pushing parallel up AB,

and tension force T pushing parallel up AB.

There are 4 forces acting on block E:

Weight force P pulling down,

Normal force N₂ pushing perpendicular to BC,

Friction force N₂μ pushing parallel to BC,

and tension force T pulling parallel to BC.

Sum of forces on D in the perpendicular direction:

∑F = ma

N₁ − P sin θ = 0

N₁ = P sin θ

Sum of forces on D in the parallel direction:

∑F = ma

T + N₁μ − P cos θ = 0

T = P cos θ − N₁μ

T = P cos θ − P sin θ μ

T = P (cos θ − sin θ μ)

Sum of forces on E in the perpendicular direction:

∑F = ma

N₂ − P cos θ = 0

N₂ = P cos θ

Sum of forces on E in the parallel direction:

∑F = ma

N₂μ + P sin θ − T = 0

T = N₂μ + P sin θ

T = P cos θ μ + P sin θ

T = P (cos θ μ + sin θ)

Set equal:

P (cos θ − sin θ μ) = P (cos θ μ + sin θ)

cos θ − sin θ μ = cos θ μ + sin θ

1 − tan θ μ = μ + tan θ

1 − μ = tan θ μ + tan θ

1 − μ = tan θ (μ + 1)

tan θ = (1 − μ) / (1 + μ)

Plug in values:

tan θ = (1 − 0.4) / (1 + 0.4)

θ = 23.2°

∠BCA = 45°, so the angle of AC relative to the horizontal is 45° − 23.2° = 21.8°.

3 0
4 years ago
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