Friction is a force that ___ or works against motion

An object of mass m travels along the parabola yequalsx squared with a constant speed of 5 units/sec. What is the force on the

Dmitriy789 [7]

Explanation:

The object is moving along the parabola y = x² and is at the point (√2, 2). Because the object is changing directions, it has a centripetal acceleration towards the center of the circle of curvature.

First, we need to find the radius of curvature. This is given by the equation:

R = [1 + (y')²]^(³/₂) / |y"|

y' = 2x and y" = 2:

R = [1 + (2x)²]^(³/₂) / |2|

R = (1 + 4x²)^(³/₂) / 2

At x = √2:

R = (1 + 4(√2)²)^(³/₂) / 2

R = (9)^(³/₂) / 2

R = 27 / 2

R = 13.5

So the centripetal force is:

F = m v² / r

F = m (5)² / 13.5

F = 1.85 m

7 0

3 years ago

Why does a stationary electromagnet attached to an AC source induce current in a wire coil?

lora16 [44]

This is basically Michael Faraday's law and this is known as electromagnetic induction

That's all I know

4 0

3 years ago

Two horizontal pipes have the same diameter, but pipe B is twice as long as pipe A. Water undergoes viscous flow in both pipes,

zheka24 [161]

Answer:

c) 2Q

Explanation:

From the given information:

The pressure inside a pipe can be expressed by using the formula:

$\Delta P = \dfrac{128 \mu L Q}{\pi D^4}$

Since the diameter in both pipes is the same, we can say:

$D = D_A = D_B$

where;

length of the first pipe A $L_A = L$ and the length of the second pipe B $L_B = 2L$

Since the difference in pressure is equivalent in both pipes:

Then:

$\dfrac{128 \mu L_1Q_1}{\pi D_1^4} = \dfrac{128 \mu L_2Q_2}{\pi D_2^4}$

$\dfrac{ L_1Q_1}{D_1^4} = \dfrac{ L_2Q_2}{D_2^4}$

$\dfrac{ LQ_1}{D^4} = \dfrac{ 2LQ}{D^4}$

$\mathbf{Q_1 = 2Q}$

6 0

3 years ago

What is one of the drawbacks to the use of solar energy for generating electricity?

-Dominant- [34]

Solar energy is one of the renewable energy sources that is invested around the world because of its large potential. the sun as a free source can sometimes bring a drawback. one of the drawbacks is that it cannot be harnessed when storms or rains are present in the weather. it is also expensive and young.

7 0

3 years ago

Pls help me i have to submit this due tomorrow :'D

aniked [119]

$\textsf {a) As there is no change in the initial and final positions, the displacement is}\\\textsf {equal to 0.}$

b) Finding total distance :

Distance travelled from 0 s to 5 s :

1/2 x 5 x 30 = 75 m

Distance travelled from 5 s to 10 s :

5 x 30 = 150 m

Distance travelled from 10 s to 15 s :

150 + 1/2 x 5 x 5
150 + 12.5 = 162.5 m

Distance travelled from 15 s to 20 s :

36 x 5 + 1/2 x 5 x 4
180 + 10 = 190 m

Distance travelled from 20 s to 25 s :

45 x 5 + 1/2 x 5 x 5
225 + 12.5 = 237.5 m

Distance travelled from 25 s to 30 s :

40 x 5 + 1/2 x 10 x 5
200 + 25 = 225 m

Distance travelled from 30 s to 35 s :

20 x 5 + 1/2 x 20 x 5
100 + 50 = 150 m

Distance travelled from 35 s to 40 s :

1/2 x 20 x 5
50 m

Total = 75 + 150 + 162.5 + 190 + 237.5 + 225 + 150 + 50

Total = 1240 m

c) velocity at t = 15 s

36/15
12/5
2.4 m/s

d) average velocity

0 m/s (as displacement is equal to 0)

e) average speed

1240/40
31 m/s

f) Part d uses displacement whereas part e uses distance

3 0

2 years ago