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4 years ago

A 2.5 kg ball rolls forward at 10.0 m/s. What is the ball's momentum?

Physics

1 answer:

Anuta_ua [19.1K]4 years ago

4 0

Answer:25kgm/s

Explanation:

mass=2.5kg

Velocity=10m/s

Momentum=mass x velocity

Momentum=2.5 x 10

Momentum=25kgm/s

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Two identical charges q placed 2.0 mapart exert forces of magnitude 4.0 N on each other What is the value of the charge q? a) q

katen-ka-za [31]

Answer:

c) $4.2*10^{-5}C$

Explanation:

Coulomb's law says that the force exerted between two charges is inversely proportional to the square of distance between them, and is given by the expression:

$F=\frac{kq_{1}q_{2}}{r^{2}}$

where k is a proportionality constant with the value $k=9*10^{9}\frac{Nm^{2}}{C^{2}}$

In this case $q_{1}=q_{2}=q$ , so we have:

$F=\frac{kq^{2}}{r^{2}}$

Solving the equation for q, we have:

$kq^{2}=Fr^{2}$

$q^{2}=\frac{Fr^{2}}{k}$

$q=\sqrt{\frac{Fr^{2}}{k}}$

Replacing the given values:

$q=\sqrt{\frac{4.0N*(2.0m)^{2}}{9*10^{-9}\frac{Nm^{2}}{C^{2}}}}$

$q=4.2*10^{-5}C$

3 0

3 years ago

Use the following terms in the same sentence:proton, neutron ,and isotope

UkoKoshka [18]

Atoms of the same element having equal numbers of protons, but different numbers of neutrons is called isotope.

8 0

3 years ago

A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop.

Lynna [10]

Answer:

-0.0047 rad/s²

335.103 seconds

99.18 seconds

Explanation:

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity = 1.5 ra/s

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation = 40 rev

t = Time taken

Equation of rotational motion

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-1.5^2}{2\times 2\pi \times 40}\\\Rightarrow \alpha=-0.0047\ rad/s^2$

Acceleration while slowing down is -0.0047 rad/s²

$t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-1.5}{-0.0047}\\\Rightarrow t=335.103\ s$

Time taken to slow down is 335.103 seconds

$\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 20\times 2\pi=1.5\times t+\frac{1}{2}\times -0.0047\times t^2\\\Rightarrow 0.00235t^2-1.5t+125.66=0$

Solving the equation

$t=\frac{-\left(-1.5\right)+\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}, \frac{-\left(-1.5\right)-\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}\\\Rightarrow t=539.11, 99.18\ s$

The time required for it to complete the first 20 is 99.18 seconds as 539.11>335.103

4 0

3 years ago

Is it possible for one ray to be shorter in length than another?

8090 [49]

Yes, even light rays can vary in wavelength and frequency, if the length of the ray is sorter, it becomes more energetic and has a higher frequency. If you're talking about a ray tracing diagram for lenses or mirrors, the length of the ray doesn't really matter unless you're finding the path length but there are some procedures for that too. Let me know if I missed what you were asking.

8 0

3 years ago

An open-end mercury manometer is connected to a low-pressure pipeline that supplies a gas to a laboratory. Because paint was spi

almond37 [142]

Answer:

A. Using

Pgauge= Pmanometer

And we know that

Pgauge= deta(hpg)

So deta h = Pgauge/density x g

= 7.5(6894.76/1psig)/ 13.6*10^3*9.8m/s²)

= 387.9mm

So to find height of pipe connected to the pipe we say

= h -deta h

= 900-387.97mm

=512.02mm

B. We use manometry principle

Pgas+density xg(25*10^3)-PX density{h-(H-0.25)=0

Finally Pgas= 6.54psig

3 0

4 years ago

A 2.5 kg ball rolls forward at 10.0 m/s. What is the ball's momentum?​

A 2.5 kg ball rolls forward at 10.0 m/s. What is the ball's momentum?