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oksian1 [2.3K]
2 years ago
7

A student throws a 130 g snowball at 6.5 m/s at the side of the schoolhouse, where it hits and sticks. What is the magnitude of

the average force on the wall if the duration of the collision is 0.18 s?
Physics
1 answer:
Alex73 [517]2 years ago
8 0

Answer:

4.7 N

Explanation:

130 g = 0.13 kg

The momentum of the snowball when it's thrown at the wall is

p = mv = 0.13*6.5 = 0.845 kgm/s

Which is also the impulse. From here we can calculate the magnitude of the average force F knowing the duration of the collision is 0.18 s

p = F\Delta t

F*0.18 = 0.845

F = 0.845 / 0.18 = 4.7 N

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1. Which object is farthest from the origin at t=2sec
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Which of the following is a consequence of the special theory of relativity?
aleksandr82 [10.1K]

Answer:

D.

Explanation:

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3 0
1 year ago
Betty weighs 400 N and she is sitting on a playground swing seat that hangs 0.21 m above the ground. Tom pulls the swing back an
disa [49]

Answer:

4.15 m/s

Explanation:

As the total energy must be conserved (neglecting air resistance) the change in gravitational potential energy, must be equal to the change in kinetic energy:

ΔE = ΔK + ΔU =0

If we take as a zero reference level for the gravitational potential energy, the height of the swing seat above the ground, (which is equal to 0.21 m), we can find the initial gravitational energy, considering the height of the point where the seat is released, regarding this point:

h₀ = 1.09 m -0.21 m = 0.88 m

⇒ U₀ = m*g*h₀ = 400 N*0.88 m = 352 J

As Uf = 0, ΔU = Uf -U₀ = -352 J

As the swing starts from rest, K₀=0, so we can say:

ΔK = Kf = \frac{1}{2} *m*vf^{2}  (1)

As ΔK = -ΔU ⇒ ΔK = 352 J (2)

From (1) and (2) we can solve for vf, as follows:

vf = \sqrt{\frac{2*352J}{40.8kg}} = 4.15 m/s

So, when the swing passes through its lowest position, Betty moves at 4.15 m/s.

5 0
3 years ago
An empty truck traveling at 10 km/h has kinetic energy. How much kinetic energy does it have when loaded so that its mass and it
Katena32 [7]

Answer:

8 time increase in K.E.

Explanation:

Consider Mass of truck = m kg and speed = v m/s then

K.E. = 1/2 ×mv²

If mass and speed both are doubled i.e let m₀ = 2m and v₀ = 2v then

(K.E.)₀ =  1/2 ×2m(2v)²

(K.E.)₀ = 8 (1/2 × mv²) = 8 × K.E.

7 0
3 years ago
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