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EleoNora [17]
3 years ago
10

An object has a mass of 13.5 kilograms. What force is required to accelerate it to a rate of 9.5 m/s2?

Physics
1 answer:
Zigmanuir [339]3 years ago
4 0
Force can be expressed as the product of mass and acceleration. Mathematically, that's F = m(a). Plugging the given into the equation, we have F = (13.5 kg)(9.5 m/s²) = 128.3 kg.m/s² or 128.3 N<span>. </span>
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8 0
3 years ago
A spring with k = 53 N/m hangs vertically next to a ruler. The end of the spring is next to the 18 cm mark on the ruler. If a 2.
Anarel [89]

Answer:

1.07 m

Explanation:

x = Compression of the spring

k = Spring constant = 53 N/m

Initial length = 18 cm

P = Kinetic energy

K = Kinetic energy

At the lowest point of the mass the energy conservation is as follows

P_{ig}+P_{is}+K_i=P_{fg}+P_{fs}+K_f\\\Rightarrow mgx+0+0=mgx+\frac{1}{2}kx^2\\\Rightarrow x=\frac{2mg}{k}\\\Rightarrow x=\frac{2\times 2.4\times 9.81}{53}\\\Rightarrow x=0.89\ m

At its lowest position the mark on the ruler will be

x_f=0.18+0.89\\\Rightarrow x_f=1.07\ m

The spring line will end up at 1.07 m

4 0
3 years ago
Andrea and the other members of the tennis team are practicing hitting tennis balls. If two people hit identical balls at the sa
Firlakuza [10]

The tennis ball that is hit the hardest would have the most kinetic energy.

8 0
3 years ago
Read 2 more answers
(a) Are the radiating electric field lines around a charged particle straight lines when the particle is stationary?
s2008m [1.1K]

Answer:

Yes

Explanation:

(a)

Yes, the radiating electric field lines around a stationary charged particle are straight lines when the particle is stationary.

7 0
4 years ago
a uniform rod of length 1.5m is placed over a wedge at 0.5m from one end .a force of 100 N is applied at its one end near the we
andreev551 [17]

Explanation:

The rod is uniform, so the center of gravity is at the center, or 0.75 m from the end.  The wedge is 0.5 m from the end, so the center is 0.25 m from the wedge.

Sum the torques about the wedge (it may help to draw a diagram first).  Take counterclockwise to be positive.

∑τ = Iα

W (0.25 m) − (100 N) (0.50 m) = 0

W = 200 N

Sum the forces in the y direction.

∑F = ma

F − 100 N − 200 N = 0

F = 300 N

8 0
3 years ago
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