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3 years ago

A body moving with uniform acceleration has

Physics

1 answer:

ANEK [815]3 years ago

4 0

The x-component of the acceleration is $-11.3 cm/s^2$

Explanation:

This problem is an example of uniformly accelerated motion, so we can solve it by using the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where

u is the initial velocity

a is the acceleration

s is the displacement

t is the time

For the body in this problem, we have:

u = 11.1 cm/s is the initial velocity

The displacement is

s = -3.58 - (3.59) = -7.17 cm

While the time is

t = 2.48 s

Substituting and solving the formula for a, we find the x-component of the acceleration:

$a=\frac{2s}{t^2}-\frac{2u}{t}=\frac{2(-7.17)}{(2.48)^2}-\frac{2(11.1)}{2.48}=-11.3 cm/s^2$

Learn more about acceleration:

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a student balances a 1.5kg of broom by placing her finger 1.4m from the end of the broom handle. How far from the broom handle w

Delicious77 [7]

The broom handle that she have to balance if she hung a 400g mass from the end of the broom handle is 5.24m

This problem is centered on moment. Moment is the turning effect of a force about a point. It is expressed as:

Moment = Force× Distance

According to principle of moment, the sum of clockwise moment is equal to sum of anticlockwise moment at shown

M1d1 = M2d2

Given the following

M1 = 1.5kg

d1 = 1.4m

M2 = 400g = 0.4kg

d2 is required

Substitute

1.5(1.4) = 0.4d2

2.1 = 0.4d2

d2 = 2.1/0.4

d2 = 5.24m

Hence the broom handle that she have to if she hung a 400g mass from the end of the broom handle is 5.24m

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3 years ago

Convert 100 Newton into dyne

Inga [223]

Answer:10000000

Explanation:

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3 years ago

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Which has the greater change in momentum, a 50 gram clay ball that strikes a wall at 1 m/s and sticks or a 50 gram superball tha

jek_recluse [69]

Answer: 50 gram superball that strikes the wall at 1 m/s and bounces away at 0.8 m/s has greater change in kinetic energy.

Explanation:

50 gram superball that strikes the wall at 1 m/s and bounces away at 0.8 m/s has the greater change in kinetic energy because the collision is elastic in nature that is bodies separates after collision and doesn't lose any kinetic energy.

Also for an elastic collision, both the momentum and energy of the bodies are conserved compare to inelastic collision where only momentum is conserved but not the kinetic energy(this is attributed to bodies that sticks together after collision).

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3 years ago

While solving a problem, Tran calculates an answer that has coulomb-volts as the units.

Bess [88]

Apparently, the question is looking for A. electric potential energy;

but I don't think that's quite right. Electric potential difference is expressed in Joules / Coulomb which is the work to move a charge between 2 points

Example: If the electric field between, say, between 2 capacitor plates is

E = 100 Newtons / Coulomb then the work done in moving a unit of charge from the negative plate to the positive plate separted by 1 cm is

V = E * d = 100 Newtons / Coulomb * .01 meters = 1 Newton-meter / Coulomb

= 1 Joule / Coulomb which is the electric potential or potential difference

(The definition of electric potential between points is "the work moving a unit positive test charge from one point to the other")

Now in our above example where V = 1 Joule / Coulomb

if we move 10 Coulombs from the negative plate to the positive plate

W = V Q = 1 Joule / Coulomb * 10 Coulombs = 10 Joules

where work done has the correct units of Joules.

Your textbook should help clarify this.

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3 years ago

Given two metal balls (that are identical) with charges LaTeX: q_1q 1and LaTeX: q_2q 2. We find a repulsive force one exerts on

Romashka [77]

Answer:

$q_1=\pm0.03 \mu C$ and $q_2=\pm0.02 \mu C.$

Explanation:

According to Coulomb's law, the magnitude of force between two point object having change $q_1$ and $q_2$ and by a dicstanced is

$F_c=\frac{1}{4\pi\spsilon_0}\frac{q_1q_2}{d^2}-\;\cdots(i)$

Where, $\epsilon_0$ is the permitivity of free space and

$\frac{1}{4\pi\spsilon_0}=9\times10^9$ in SI unit.

Before dcollision:

Charges on both the sphere are $q_1$ and $q_2$ , d=20cm=0.2m, and $F_c=1.35\times10^{-4}$ N

So, from equation (i)

$1.35\times10^{-4}=9\times10^9\frac{q_1q_2}{(0.2)^2}$

$\Rightarrow q_1q_2=6\times10^{-16}\;\cdots(ii)$

After dcollision: Each ephere have same charge, as at the time of collision there was contach and due to this charge get redistributed which made the charge density equal for both the sphere t. So, both have equal amount of charhe as both are identical.

Charges on both the sphere are mean of total charge, i.e

$\frac{q_1+q_2}{2}$

d=20cm=0.2m, and $F_c=1.406\times10^{-4}$ N

So, from equation (i)

$1.406\times10^{-4}=9\times10^9\frac{\left(\frac{q_1+q_2}{2}\right)^2}{(0.2)^2}$

$\Rightarrow (q_1+q_2)^2=2.50\times10^{-15}$

$\Rightarrow q_1+q_2=\pm5\times 10^{-8}$

As given that the force is repulsive, so both the sphere have the same nature of charge, either positive or negative, so, here take the magnitude of the charge.

$\Rightarrow q_1+q_2=5\times 10^{-8}\;\cdots(iii)$