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2 years ago

How does collision time affect impact forces

Physics

1 answer:

Licemer1 [7]2 years ago

5 0

Answer:

A large increase in collision time can mean decrease in impact force. This is essentially what an air bag in your car does. It increases the time over which it stops you and decreases the force. ... In this case, the piano would have a smaller change in momentum and a smaller impact force.

Explanation:

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By how much does the pressure of a gas in a rigid vessel decrease when the temperature drops from 0 ∘C to –1∘C?

Ugo [173]

When a problem says a rigid vessel, it means that volume is constant. At constant V, pressure and temperature are indirectly proportional. We calculate as follows:

P1/T1 = P2/T2
P1/P2 = T1/T2
P1/P2 = 273.15 / 272.15
P1/P2 = 1.00

Hope this helps. Have a nice day.

7 0

3 years ago

Objects 1 and 2 attract each other with a gravitational force of 12 units. If the mass of Object 2 is tripled, then the new grav

olasank [31]

Explanation:

Fgravity = G*(mass1*mass2)/D².

G is the gravitational constant, which has the same value throughout our universe.

D is the distance between the objects.

so, if you triple one of the masses, what does that do to our equation ?

Fgravitynew = G*(3*mass1*mass2)/D²

due to the commutative property of multiplication

Fgravitynew = 3* G*(mass1*mass2)/D² = 3* Fgravity

so, the right answer is 3×12 = 36 units.

5 0

1 year ago

One ring of radius a is uniformly charged with charge +Q and is placed so its axis is the x-axis. A second ring with charge –Q i

kati45 [8]

Answer:

The force exerted on an electron is $7.2\times10^{-18}\ N$

Explanation:

Given that,

Charge = 3 μC

Radius a=1 m

Distance = 5 m

We need to calculate the electric field at any point on the axis of a charged ring

Using formula of electric field

$E=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

$E_{1}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times3\times10^{-6}\times5}{(1^2+5^2)^{\frac{3}{2}}}$

$E_{1}=1.0183\times10^{3}\ N/C$

Using formula of electric field again

$E_{2}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times(-3\times10^{-6})\times5}{((0.5)^2+5^2)^{\frac{3}{2}}}$

$E_{2}=-1.064\times10^{3}\ N/C$

We need to calculate the resultant electric field

Using formula of electric field

$E=E_{1}+E_{2}$

Put the value into the formula

$E=1.0183\times10^{3}-1.064\times10^{3}$

$E=-0.045\times10^{3}\ N/C$

We need to calculate the force exerted on an electron

Using formula of electric field

$E = \dfrac{F}{q}$

$F=E\times q$

Put the value into the formula

$F=-0.045\times10^{3}\times(-1.6\times10^{-19})$

$F=7.2\times10^{-18}\ N$

Hence, The force exerted on an electron is $7.2\times10^{-18}\ N$

8 0

2 years ago

The combined-gas law relates which of these?

Fofino [41]

The combined-gas law relates which temperature, pressure and volume.

Temperature=T
Pressure=P
Volume=V

(P₁*V₁) / T₁=(P₂*V₂) / T₂

D. Temperature, pressuere and volume.

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2 years ago

If the energy from the light is captured and stored by the surface without being reflected or transmitted, then it has been ____

vekshin1

A. ABSORBEDDDDDDDDDDDDDDDDDD

6 0

3 years ago

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