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ioda
3 years ago
15

How to calculate average breaking force physics

Physics
1 answer:
Alexandra [31]3 years ago
8 0

Answer:

Formula: Remove the zero from the speed, multiply the figure by itself and then multiply by 0.4. The figure 0.4 is taken from the fact that the braking distance from 10 km/h in dry road conditions is approximately 0.4 metres.

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A force in the negative direction of an x-axis is applied for 17ms to a 0.18 kg ball initially moving at 16.0 m/s in the positiv
Marizza181 [45]
Impulse = Ft=mΔv => Δv = Ft/m = 4.28/0.18 = 23.78 m/s

But,
Δv = v1-v2, where v1 = initial velocity = 16 m/s, v2 = final velocity

Therefore,
v1 - v2 = 23.78 => v2 = v1 - 23.78 => v2 = 16 - 23.78 = -7.78 m/s

The velocity of ball after the force is 7.78 m/s in the direction of the force.

5 0
3 years ago
C is correct just an FYI
belka [17]

Answer:

Huh?

Explanation:

3 0
2 years ago
An astronaut is standing on the surface of a planetary satellite that has a radius of 1.74 times 10^6 m and a mass of 7.35 times
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Answer:

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3 years ago
Determine the velocity of the running man with respect to car A.
olasank [31]

Answer:

A) -55  km/

hr

Explanation:

3 0
3 years ago
Read 2 more answers
A coil is wrapped with 300 turns of wire on the perimeter of a circular frame (radius = 8.0 cm). Each turn has the same area, eq
TiliK225 [7]

Answer:

Approximately 18 volts when the magnetic field strength increases from \rm 20\; mT to \rm 80\;mT at a constant rate.

Explanation:

By the Faraday's Law of Induction, the EMF \epsilon that a changing magnetic flux induces in a coil is:

\displaystyle \epsilon = N \cdot \frac{d\phi}{dt},

where

  • N is the number of turns in the coil, and
  • \displaystyle \frac{d\phi}{dt} is the rate of change in magnetic flux through this coil.

However, for a coil the magnetic flux \phi is equal to

\phi = B \cdot A\cdot \cos{\theta},

where

  • B is the magnetic field strength at the coil, and
  • A\cdot \cos{\theta} is the area of the coil perpendicular to the magnetic field.

For this coil, the magnetic field is perpendicular to coil, so \theta = 0 and A\cdot \cos{\theta} = A. The area of this circular coil is equal to \pi\cdot r^{2} = \pi\times 8.0\times 10^{-2}\approx \rm 0.0201062\; m^{2}.

A\cdot \cos{\theta} = A doesn't change, so the rate of change in the magnetic flux \phi through the coil depends only on the rate of change in the magnetic field strength B. The size of the magnetic field at the instant that B = \rm 50\; mT will not matter as long as the rate of change in B is constant.

\displaystyle \begin{aligned} \frac{d\phi}{dt} &= \frac{\Delta B}{\Delta t}\times A \\&= \rm \frac{80\times 10^{-3}\; T- 20\times 10^{-3}\; T}{20\times 10^{-3}\; s}\times 0.0201062\;m^{2}\\&= \rm 0.0603186\; T\cdot m^{2}\cdot s^{-1}\end{aligned}.

As a result,

\displaystyle \epsilon = N \cdot \frac{d\phi}{dt} = \rm 300 \times 0.0603186\; T\cdot m^{2}\cdot s^{-1} \approx 18\; V.

6 0
3 years ago
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