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storchak [24]
3 years ago
8

A student drops a ball off the top of building and records that the ball takes 3.32s to reach the ground (g = 9.8 m/s^2). What i

s the ball speed just before hitting the ground?​
Physics
1 answer:
slega [8]3 years ago
7 0

Answer:

Explanation:

Here's what we know because it was given to us:

a = -9.8 m/s/s and

time = 3.32 seconds

Here's what we know because we rock physics:

v₀ = 0 (because the object was held still before it was dropped).

Here's the equation that ties all that info together in a single one-dimensional equation:

v = v₀ + at

Filling in and solving for v:

v = 0 + (-9.8)(3.32) and

v = -33m/s

The velocity is negative because the object is moving downwards and up is positive (but you knew that already too!)

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What are the effects of ultraviolet sun rays????​
pychu [463]

Answer: UV rays, either from the sun or from artificial sources like tanning beds, can cause sunburn. Exposure to UV rays can cause premature aging of the skin and signs of sun damage such as wrinkles, leathery skin, liver spots, actinic keratosis, and solar elastosis. UV rays can also cause eye problems.

Explanation: Uv means ultraviolet

3 0
3 years ago
Plzpzlpzlzplzplzplzpz this one also<br> all questions ​
WINSTONCH [101]

Answer:

I didn't know these questions sorry

4 0
3 years ago
To practice problem-solving strategy 22.1: gauss's law. an infinite cylindrical rod has a uniform volume charge density ρ (where
BabaBlast [244]

Let say the point is inside the cylinder

then as per Gauss' law we have

\int E.dA = \frac{q}{\epcilon_0}

here q = charge inside the gaussian surface.

Now if our point is inside the cylinder then we can say that gaussian surface has charge less than total charge.

we will calculate the charge first which is given as

q = \int \rho dV

q = \rho * \pi r^2 *L

now using the equation of Gauss law we will have

\int E.dA = \frac{\rho * \pi r^2* L}{\epcilon_0}

E. 2\pi r L = \frac{\rho * \pi r^2* L}{\epcilon_0}

now we will have

E = \frac{\rho r}{2 \epcilon_0}

Now if we have a situation that the point lies outside the cylinder

we will calculate the charge first which is given as it is now the total charge of the cylinder

q = \int \rho dV

q = \rho * \pi r_0^2 *L

now using the equation of Gauss law we will have

\int E.dA = \frac{\rho * \pi r_0^2* L}{\epcilon_0}

E. 2\pi r L = \frac{\rho * \pi r_0^2* L}{\epcilon_0}

now we will have

E = \frac{\rho r_0^2}{2 \epcilon_0 r}


7 0
3 years ago
You have a grindstone (a disk) that is 90.0 kg, has a 0.340-m radius, and is turning at 90.0 rpm, and you press a steel axe agai
QveST [7]

Answer:

Explanation:

Given mass of grindstone m=90\ kg

radius of stone r=0.34\ m

angular speed of disc \omega =90\ rpm

Steel Axle applying a force of F=20\ N

coefficient of kinetic friction \mu =0.2

Frictional Torque applied by steel is given by

\tau=r\times f_r

\tau =r\times \mu F

where f_r=frictional force

\tau =r\times \mu \times F

\tau =0.34\times 0.2\times 20

\tau =1.36\ N-m

Torque is also given by

\tau =I\cdot \alpha

where \alpha=angular acceleration

I=moment of Inertia

\tau =0.5Mr^2\times \alpha

0.5Mr^2\times \alpha =1.36\ N-m

0.5\times 90\times 0.34^2\times \alpha =1.36

\alpha =0.261\ rad/s^2

3 0
4 years ago
You are pushing a cart across the room, and the cart has wheels at the front and the back. Your hands are placed on top of the c
Marizza181 [45]

Answer:

The normal force is the force that the floor does as a reaction of the gravitational force that an object does against the floor (is the resistance that objects have when other objects want to move trhough them, and the force comes by the 3rd Newton's law, and this is specially used in cases where the first object is fixed, like walls or the floor). With this in mind, the point in where the normal force will be greater is the point that is closer to the center of mass of the object (the point with more mass)

If the wheels are in the extremes of the object, and the center of mass is in the middle of the object, the normal force will be equal. Now if for example, you put a little mass in one end of the object, now the center of weight displaces a little bit and is not centered, and the side is where you put the weight on will receive a bigger normal force from the floor than the other side.

6 0
3 years ago
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