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yarga [219]
3 years ago
14

2. A rocket is launched with an upward acceleration of 20.0 m/s2 that lasts for 4.50 s. During

Physics
1 answer:
Neporo4naja [7]3 years ago
8 0

Answer:

Ggghghhh

Explanation:

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In a crash test, a 2,500 kg car hits a concrete barrier at 13 m/s2 calculate the amount of force at which the car strikes the ba
Marta_Voda [28]

Answer:

32500N

Explanation:

Data obtained from the question include:

m (mass) = 2500 kg

a (acceleration) = 13 m/s2

F (force) =?

Force is the product of mass and acceleration. It is represented mathematically as:

Force = mass x acceleration

F = m x a

With the above formula, the force with which the car strikes the barrier can be obtained as follow:

F = m x a

F = 2500 x 13

F = 32500N

Therefore, the car will strike the barrier with a force of 32500N

8 0
3 years ago
Read 2 more answers
What force is needed to accelerate 300kg rock as a rste of 4m/s​
Natali [406]

Answer:

60 maybe

Explanation:

4 0
3 years ago
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a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance
aivan3 [116]

Answer:

(a) Electric field at 0.250 m is zero.

(b)  Electric field at 2.90 m is zero.

(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.

(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.

Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

  = \frac{KQ}{r^{2} } for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b)  Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

(c) Electric field at 3.10 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{3.10^{2} }

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{8^{2} }

E = - 0.77 x 10³ V/m

8 0
3 years ago
In Paul Hewitt's book, he poses this question: "If the forces that act on a bullet and the recoiling gun from which it is fired
Sauron [17]
They have different accelerations because of their masses. According to Newton's Second Law, an objects acceleration is inversely proportional to its mass. Therefore the object with the larger mass, in this case the gun, will have a smaller acceleration. In the same way, the less massive object, being the bullet, will have a higher acceleration.

Hope this helps :)
4 0
3 years ago
At what speed do a bicycle and its rider, with a combined mass of 100 kg, have the same momentum as a 1400 kg car traveling at 2
likoan [24]
100kg x bicycle speed = 1400 X 2

bicycle speed = 2800/ 100
bicycle speed = 28 m/s
3 0
4 years ago
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