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3 years ago

9

a net force of 25 newtons is applied horizontally to 10. -kilogram block resting on a table. what is the magnitude of the accele

ration of the block?

1 answer:

Pavel [41]3 years ago

4 0

Answer:2.5m/s^2

Explanation:

Acceleration=force/mass

A=25/10

A=2.5

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390.625
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3 years ago

Explain the difference between radiation as it is used in most everyday language and radiation as it is used in an astronomical

horsena [70]

Answer:Explained Below

Explanation:

Radiation in everyday language refers to certain type of subatomic particle released by a radioactive unstable nucleus.It can be used to kill the cancer by damaging the DNA of cancer cells.

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3 years ago

What is the maximum value of the electric field in an electromagnetic wave whose average intensity is 4.50 w/m2 ?

Oksi-84 [34.3K]

Average intensity of electromagnetic wave is given by

I= (speed of light*permittivity*sq. electric field)/2

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Permittivity = 8.85x10^-12

Therefore,
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3 years ago

A 800-gram grinding wheel 27.0 cm in diameter is in the shape of a uniform solid disk. (We can ignore the small hole at the cent

SSSSS [86.1K]

Explanation:

d = Diameter of wheel = 27 cm

r = Radius = $\frac{d}{2}=\frac{27}{2}=13.5\ cm$

m = Mass of wheel = 800 g

$\omega_i$ = Initial angular velocity = $245\times \frac{2\pi}{60}\ rad/s$

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-245\times \frac{2\pi}{60}}{50}\\\Rightarrow \alpha=-0.51312\ rad/s^2$

Moment of inertia is given by

$M=\frac{1}{2}mr^2\\\Rightarrow M=\frac{1}{2}\times 0.8\times 0.135^2\\\Rightarrow M=0.00729\ kgm^2$

Torque is given by

$\tau=I\alpha\\\Rightarrow \tau=0.00729\times -0.51312\\\Rightarrow \tau=-0.0037406448\ Nm$

The torque the friction exerts is -0.0037406448 Nm

For more information on torque and moment of inertia refer

brainly.com/question/13936874

brainly.com/question/3406242

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3 years ago

Ask Your Teacher Cam Newton of the Carolina Panthers throws a perfect football spiral at 6.9 rev/s. The radius of a pro football

faltersainse [42]

Answer:

$a=159.32\ m/s^2$

Explanation:

It is given that,

Angular speed of the football spiral, $\omega=6.9\ rev/s=43.35\ rad/s$

Radius of a pro football, r = 8.5 cm = 0.085 m

The velocity is given by :

$v=r\omega$

$v=0.085\times 43.35$

v = 3.68 m/s

The centripetal acceleration is given by :

$a=\dfrac{v^2}{r}$

$a=\dfrac{(3.68)^2}{0.085}$

$a=159.32\ m/s^2$

So, the centripetal acceleration of the laces on the football is $159.32\ m/s^2$ . Hence, this is the required solution.

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3 years ago