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Lana71 [14]
3 years ago
5

When the mover pushes the box, two equal forces result. Explain why the box moves

Physics
1 answer:
pychu [463]3 years ago
7 0
The mover is pushing on the box, meaning that when hes pushing on the box hes adding force to the box to make it move. The box cant move without force.
You might be interested in
How much mass should be attached to a vertical ideal spring having a spring constant (force constant) of 39.5 n/m so that it wil
mrs_skeptik [129]
The frequency of a simple harmonic oscillator such as a spring-mass system is given by
f= \frac{1}{2 \pi}   \sqrt{ \frac{k}{m} }
where 
k is the spring constant
m is the mass attached to the spring.

Re-arranging the formula, we get:
m= \frac{k}{4 \pi^2 f^2}
and since we know the constant of the spring:
k=39.5 N/m
and the frequency of oscillation:
f=1.00 Hz
we can find the value of the mass attached to it:
m= \frac{39.5 Hz}{4 \pi^2 (1.00 Hz)^2} = 1.00 kg
7 0
3 years ago
Three kids are riding on a snow sled traveling horizontally without friction at 19.8 m/s. The masses of Kid A, B, and C are 42.8
Kamila [148]

Answer:

34.6 m/s

Explanation:

From conservation of momentum, the sum of initial and final momentum are equal. Momentum is a product of mass and velocity. Initial mass will be 42.8+31.5+25.9=100.2 kg

Final mass will be 31.5+25.9=57.4 kg

From formula of momentum

M1v1=m2v2

Making v2 the subject of the formula then

V2=\frac {M1v1}{m2}

Substitute 100.2 kg for M1, 19.8 m/s fkr v1 and 57.4 kg for m2 then

V2=\frac {100.2 kg\times 19.8 m/s}{57.4 kg}=34.56376 m/s\approx 34.6 m/s

4 0
3 years ago
A hippo drives 42 km due East. He then turns and drives 28 km at 25° East of South. He turns again and drives 32 km at 40° North
ch4aika [34]

Answer:

a) Please, see the attched figure

b) Total displacement R = (78.3 km; -4.8 km)

c) R = (78.4 km * cos (-3.5°); 78.4 km * sin (-3.5°))

d) The hippo is 78.4 km from his starting point.

The total distance traveled is 102 km

Explanation:

a)Please, see the attached figure.

b) The vector A can be expressed as:

A = (magnitude * cos α; magnitude * sin α)

Where

magnitude = 42 km

α= 0

Then,

A = (42 km ; 0) or 42 km i

In the same way, we can proceed with the other vectors:

B = ( Bx ; By)

where

(apply trigonometry of right triangles: sen α = opposite / hypotenuse and

cos α = adjacent / hypotenuse. See the figure to determine which component of vector B is the opposite and adjacent side to α)

Bx = 28 km * sin 25 = 11.8 km

By = 28 km * cos 25 = -25.4 km (it has to be negative since it is directed towards the negative vertical region according to our reference system)

B = (11.8 km; -25.4 km) or 11.8 km i - 25.4 km j

C = (Cx; Cy)

where

Cx = 32 km * cos 40° = 24.5 km

Cy = 32 km * sin 40 = 20.6 km

C = (24.5 km; 20.6 km)

Then:

R = A+B+C = (42 km + 11.8 km + 24.5 km; 0 - 25.4 km + 20.6 km)

= (78.3 km; -4.8 km) or 78.3 km i -4.8 km j

c) R = (78.3 km; -4.8 km)

The magnitude of R is:

magnitude = \sqrt{(78.3)^{2 }+ (-4.8)^{2}}= 78.4 km

Using trigonometry, we can calculate the angle:

Knowing that

tan α = opposite / adjacent

and that

opposite = Ry = -4.8 km

adjacent = Rx = 78.3 km

Then:

tan α = -4.8 km / 78.4 km

α = -3.5°

We can now write the vector R in magnitude and direction form:

R = (78.4 km * cos (-3.5°); 78.4 km * sin (-3.5°))

d) The displacement of the hipo relative to the starting point is the magnitude of vector R calculated in c):

magnitude R = 78. 4 km

The total distance traveled is the sum of the magnitudes of each vector:

Total distance = 42 km + 28 km + 32 km = 102 km  

3 0
3 years ago
Earth's gravitational pull just got threetimes stronger! what happens to your weight/
OLga [1]
Hold on and let's discuss this realistically.

Because of gravity, there are two forces between the Earth and me. One draws me toward the Earth. The strength of that force is what I call my "weight". The other force draws the Earth toward me, and has the same strength.

The strength of these forces depends on the masses of the Earth and me. If the strength just tripled, that means that at least one of us just picked up a lot more mass. If the Earth suddenly became three times as massive, then the weight of everything and everybody on it would suddenly triple, and I'm pretty sure it would be the end of all of us before too long.

If it was only MY mass that suddenly tripled, that would mean that I had gone tearing through my house and the neighbour's house, eating everything in sight including the 2 couches, 3 dogs, and 6 TVs. Naturally, just as you would expect, my weight changed from 207 to 621, and my skin is stretched really tight.
ooohhh
8 0
3 years ago
The masses are m1 = m, with initial velocity 2v0, and m2 = 7.4m, with initial velocity v0. Due to the collision, they stick toge
lesya [120]

Answer:

Loss, \Delta E=-10.63\ J

Explanation:

Given that,

Mass of particle 1, m_1=m =0.66\ kg

Mass of particle 2, m_2=7.4m =4.884\ kg

Speed of particle 1, v_1=2v_o=2\times 6=12\ m/s

Speed of particle 2, v_2=v_o=6\ m/s

To find,

The magnitude of the loss in kinetic energy after the collision.

Solve,

Two particles stick together in case of inelastic collision. Due to this, some of the kinetic energy gets lost.

Applying the conservation of momentum to find the speed of two particles after the collision.

m_1v_1+m_2v_2=(m_1+m_2)V

V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}

V=\dfrac{0.66\times 12+4.884\times 6}{(0.66+4.884)}

V = 6.71 m/s

Initial kinetic energy before the collision,

K_i=\dfrac{1}{2}(m_1v_1^2+m_2v_2^2)

K_i=\dfrac{1}{2}(0.66\times 12^2+4.884\times 6^2)

K_i=135.43\ J

Final kinetic energy after the collision,

K_f=\dfrac{1}{2}(m_1+m_2)V^2

K_f=\dfrac{1}{2}(0.66+4.884)\times 6.71^2

K_f=124.80\ J

Lost in kinetic energy,

\Delta K=K_f-K_i

\Delta K=124.80-135.43

\Delta E=-10.63\ J

Therefore, the magnitude of the loss in kinetic energy after the collision is 10.63 Joules.

7 0
3 years ago
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