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sammy [17]
2 years ago
13

A student throws a rock horizontally from the edge of a cliff that is 20 m high. The rock has an initial speed of 10 m/s. If air

resistance is negligible, the distance from the base of the cliff to where the rock hits the level ground below the cliff is most nearly
a.5m
b.10m
c.20m
d.40m
e.200m
Physics
1 answer:
scZoUnD [109]2 years ago
5 0

Answer:

c.20

Explanation:

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A. List at least 5 examples of events or factors that could cause extinction. 
baherus [9]
Disease
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8 0
3 years ago
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A gas in a piston–cylinder assembly undergoes a compression process for which the relation between pressure and volume is given
hichkok12 [17]

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Explanation:

3 0
3 years ago
A car maintains a constant speed v as it traverses the hill and valley as shown below. Both the hill and valley have a radius of
Zigmanuir [339]

Answer:

As given that the car maintains a constant speed v as it traverses the hill and valley where both the valley and hill have a radius of curvature R.

(i) At point C, the normal force acting on the car is largest because the centripetal force is up. gravity is down and normal force is up. net force is up so magnitude of normal force must be greater than the car's weight.

(ii) At point A, the normal force acting on the car is smallest because the centripetal force is down. gravity is down and normal force is up. net force is up so magnitude of normal force must be less than car's weight.

(iii) At point C, the driver will feel heaviest because the driver's apparent weight is the normal force on her body.

(iv) At point A, the driver will feel the lightest.

(v)The car can go that much fast without losing contact with the road at A can be determined as follow:

Fn=0 - lose contact with road

Fg= mv²/r

mg=mv²/r

v=sqrt (gr)

8 0
2 years ago
When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The mass is now displ
Ray Of Light [21]

Answer:

period of oscillations is 0.695 second

Explanation:

given data

mass m = 0.350 kg

spring stretches x = 12 cm = 0.12 m

to find out

period of oscillations

solution

we know here that force

force = k × x   .........1

so force = mg =  0.35 (9.8)  = 3.43 N

3.43 = k × 0.12

k = 28.58 N/m

so period of oscillations is

period of oscillations = 2π × \sqrt{\frac{m}{k} }   ................2

put here value

period of oscillations = 2π × \sqrt{\frac{0.35}{28.58} }  

period of oscillations = 0.6953

so period of oscillations is 0.695 second

4 0
3 years ago
A softball player moving 3.89 m/s
Ahat [919]

Answer:

0.119 s

Explanation:

Given that

U=3.89\ m/s\\a=-1.44\ m/s^2\\S=4.8\ m

Lets take final speed of the softball after covering 4.8 m = V

We know that

V^2=U^2+2aS\\V^2=3.89^2-2\times 1.44\times 4.8\\V=3.7\ m/s

Also We know that

V=U+at\\

Putting the value of V ,U and a in the previous equation  We get

3.7=3.89-1.44\times t\\t=0.119\ s

Therefore slide time will be 0.119 s

3 0
3 years ago
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