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DanielleElmas [232]
4 years ago
14

Which statements describe friction? Check all that apply.

Physics
2 answers:
Ad libitum [116K]4 years ago
7 0
The answer is 3, 5 and 2
svp [43]4 years ago
6 0

Answer:

The answers are 2, 3 and 5

Explanation:

Frictional force is a contact force that occurs as two bodies move or slides over each other causing a reduction in efficiency and can be reduced by greasing.

2. It causes a change in motion.

3. It occurs in the same direction as the push force.

5. It is parallel to the surfaces that rub together

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Suppose a rocket launches with an acceleration of 34.0 m/s2 what is the apparent weight of an 85-kg astronaut aboard this rocket
docker41 [41]

Answer:

the apparent weight of the astronaut in the rocket is 3723 N.

Given:

acceleration = 34 \frac{m}{s^{2} }

mass of astronaut = 85 kg

To find:

Apparent weight of the astronaut = ?

Solution:

Total weight of the astronaut in a rocket is given by,

W = w + F

W = apparent weight of the astronaut

w = weight of astronaut on earth surface = mg

F = force acting on the astronaut = ma

W = mg+ma

W = m (g+a)

W = 85 (9.8 + 34)

W = 3723 N

Thus, the apparent weight of the astronaut in the rocket is 3723 N.

7 0
3 years ago
Read 2 more answers
Please help wil give brainiest & 40p.
dexar [7]

Answer:

This is the answer

Explanation:

You can find the ans in the photo I attached.

3 0
3 years ago
A horizontal insulating rod of length 11.8-cm and charge 19 nC is in a plane with a long straight vertical uniform line charge.
SCORPION-xisa [38]

Answer:

11.962337 × 10^-4 N

Explanation:

Given the following :

Length L = 11.8

Charge = 29nC = 29 × 10^-9 C

Linear charge density λ = 1.4 × 10^-7 C/m

Radius (r) = 2cm = 2/100 = 0.02 m

Using the relation:

E = 2kλ/r ; F =qE

F = 2kλq/L × ∫dr/r

F = 2*k*q*λ/L × (In(0.02 + L) - In(0.02))

2*k*q*λ/L = [2 × (9 * 10^9) * (29 * 10^9) * (1.4 * 10^-7)]/ 0.118] = 6193.2203 × 10^(9 - 9 - 7) = 6193.2203 × 10^-7 = 6.1932203 × 10^-4

In(0.02 + 0.118) - In(0.02) = In(0.138) - In(0.02) = 1.9315214

Hence,

(6.1932203 × 10^-4) × 1.9315214 = 11.962337 × 10^-4 N

3 0
4 years ago
A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.
kow [346]

Before the engines fail, the rocket's altitude at time <em>t</em> is given by

y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

and its velocity is

v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t

The rocket then reaches an altitude of 1150 m at time <em>t</em> such that

1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

Solve for <em>t</em> to find this time to be

t=11.2\,\mathrm s

At this time, the rocket attains a velocity of

v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}

When it's in freefall, the rocket's altitude is given by

y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2

where g=9.80\frac{\rm m}{\mathrm s^2} is the acceleration due to gravity, and its velocity is

v_2(t)=124\dfrac{\rm m}{\rm s}-gt

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for y_2(t) to reach 0:

1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

{v_f}^2-{v_i}^2=2a\Delta y

where v_f and v_i denote final and initial velocities, respecitively, a denotes acceleration, and \Delta y the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means y_2 will contain the information we need to find the maximum height.

-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)

Solve for y_{\rm max} and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to y_2(t)) to be about 32.6 s. Plug this into v_2(t) to find the velocity before it crashes:

v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

3 0
4 years ago
What are 3 pros and cons of nuclear transmutation?
Tanya [424]
Pros -

It doesn't cost much
Power stations are very compact
Transportation of the material is easy

Cons-

It's not safe
Nuclear waste take 200 years to degrade
Greater risk of explosion



      -IF THIS HELPED WHICH I HOPE COMMENT BELOWw ↓PLEASEX . THANKs!
6 0
4 years ago
Read 2 more answers
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