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lys-0071 [83]
3 years ago
14

What is an example of static friction

Physics
1 answer:
dolphi86 [110]3 years ago
7 0

imagine something setting stil, such as a box, since it is not moving, it has static friction

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Explain the meaning of the name "black hole". This should be 2-3 sentences. Hint- relate this to density and how it effects thin
Rina8888 [55]
A black hole is a very dense, massive, body, left behind after a large star explodes. It has a huge gravity which will attract everything that comes near it, even light!
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3 years ago
What forces are acting on your body during a bungee jump?
LiRa [457]

Answer:

The first force that the bungee jumper experiences is gravity, which pulls down on everything and makes the jumper fall. The gravitational force is almost exactly constant throughout the jump. During the bungee jumper's fall, he or she also experiences a force due to air resistance.

Explanation:

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3 years ago
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Complete all four parts. 15 points. Will give brainliest! Show work!
Vlada [557]

Answer:

A. 5.08 secs.

B. 10.16 secs.

C. 126.50 m.

D. 373.36 m

Explanation:

Data obtained from the question include the following:

Initial velocity (u) = 65 m/s

Angle of projection θ = 50°

A. Determination of the time taken to reach the peak.

Initial velocity (u) = 65 m/s

Angle of projection θ = 50°

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =.?

t = u•Sine θ/g

t = (65 × Sine 50) /9.8

t = 5.08 secs.

B. Determination of the total time spent by the ball in air.

Time (t) taken to reach the peak = 5.08 secs.

Total time (T) spent by the ball in air =?

T = 2t

T = 2 × 5.08

T = 10.16 secs

Therefore, the total time spent by the ball in air is 10.16 secs.

C. Determination of the maximum height.

Initial velocity (u) = 65 m/s

Angle of projection θ = 50°

Acceleration due to gravity (g) = 9.8 m/s²

Maximum height (H) =..?

H = u²•Sine² θ / 2g

H = 65² × (Sine 50)² / 2 × 9.8

H = 4225 × (Sine 50)² /19.6

H = 126.50 m

Therefore, the maximum height reached by the ball is 126.50 m.

D. Determination of the horizontal distance travelled by the ball.

Initial velocity (u) = 65 m/s

Angle of projection θ = 50°

Acceleration due to gravity (g) = 9.8 m/s²

Horizontal distance (R) =..?

R = u²•Sine 2θ / g

R = 65² × Sine (2×30) / 9.8

R = (4225 × Sine 60) / 9.8

R = 373.36 m

Therefore, the horizontal distance travelled by the ball is 373.36 m

6 0
3 years ago
The magnitude of a vector can be positive or negative The magnitude of a vector can be less than the magnitude of one of its com
NARA [144]

Answer:

If magnitude of vector A is less than magnitude of vector B then its x-component will be also less      

Explanation:

In first case magnitude can never be negative -ve sign shows the direction only so it is incorrect statement.

The magnitude of any vector can never be less than its component because component of vector is found with the multiplication of sine and cos angle which values lies between 0 and 1 so after multiplication it gives less value than the vector

If magnitude of any vector is less then its x component will also be less because magnitude of component is dependent on the magnitude of vector

8 0
3 years ago
A diverging lens has a focal length of magnitude 22.8 cm. (a) Locate the images for each of the following object distances. 45.6
Nostrana [21]

(a) -15.2 cm

We can solve the problem by using the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f = -22.8 cm is the focal length of the lens (negative because it is a diverging lens)

p = 45.6 cm is the distance of the object from the lens

q is the distance of the image from the lens

Solving the equation for q, we find the position of the image:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{-22.8 cm}-\frac{1}{45.6 cm}=-0.066 cm^{-1}

q=\frac{1}{-0.066 cm^{-1}}=-15.2 cm

and the negative sign means that the image is virtual.

(b) -11.4 cm

In this case, the distance of the object from the lens is

p = 22.8 cm

Substituting into the lens equation, we can find the new image distance, q:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{-22.8 cm}-\frac{1}{22.8 cm}=-\frac{2}{22.8 cm}

q=\frac{-22.8 cm}{2}=-11.4 cm

and the negative sign means that the image is virtual.

(c) -7.6 cm

In this case, the distance of the object from the lens is

p = 11.4 cm

Substituting into the lens equation, we can find the new image distance, q:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{-22.8 cm}-\frac{1}{11.4 cm}=-0.132 cm^{-1}

q=\frac{1}{0.132 cm^{-1}}=-7.6 cm

and again, the negative sign means that the image is virtual.

3 0
3 years ago
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