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Leona [35]
3 years ago
7

A student wants to determine the impulse delivered to the lab cart when it runs into the wall. The student measures the mass of

the cart and the velocity of the cart before it hits the wall.
What other measurement is necessary to calculate the impulse delivered to the lab cart?

the velocity of the cart after it hits the wall
the force exerted on the cart by the wall
the time the cart is in contact with the wall
the acceleration of the cart before it hits the wall
Physics
1 answer:
forsale [732]3 years ago
6 0
Impulse = Force * times and also Impulse = change in momentum.

Given that the mass does not change, change if momentum = mass * (final velocity -  initial velocity)

Given that you know mass and initial velocity (which is the velicity before the cart hits the wall) you need the final velocity (which is the velocity after the cart hits the wall).

Answer: the velocity of the cart after it hits the wall.
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Increase, same, same

Explanation:

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2 years ago
Two blocks connected by a rope of negligible mass are being dragged by a horizontal force (see figure below). Suppose F = 65.0 N
vovangra [49]
Refer to the diagram shown below.

g = 9.8 m/s², and air resistance is ignored.

For mass m₁:
The normal reaction is m₁g.
The resisting force is R₁ = μm₁g.

For mass m₂:
The normal reaction is m₂g.
The resisting force is R₂ = μm₂g.

Let a =  the acceleration of the system.
Then
(m₁ + m₂)a = F - (R₁ + R₂)
(14+26 kg)*(a m/s²) = (65 N) - 0.098*(9.8 m/s²)*(14+26 kg)
40a = 65 - 38.416 = 26.584
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Answer:  0.665 m/s²  (nearest thousandth)

7 0
3 years ago
How do observing and inferring differ?
Illusion [34]
Observing is commenting on what you see e.g: It is raining.
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6 0
2 years ago
Read 2 more answers
A day on a distant planet observed orbiting a nearby star is 21.5 hr. Also, a year on the planet lasts 69.3 Earth days. In other
serg [7]

Answer:

Part A

The angular speed of rotation of the plane is 8.11781 × 10⁻⁵ rad/s

Part B

The angular speed of orbit of the planet is 1.04938 × 10⁻⁶ rad/s

Explanation:

The parameters of the planet are;

The duration of a day on the distant planet = 21.5 hr.

The duration of a year on the distant planet = 69.3 Earth days

Part A

The duration of a day = The time to make one complete revolution of 2·π radians

∴ The average angular speed about its axis, \omega_{rotation} = Angle turned/Time

∴ \omega_{rotation}  = 2·π/(21.5 × 60 × 60) s ≈ 8.11781 × 10⁻⁵ rad/s

The average angular speed of the planet about its own axis, \omega_{rotation}  = 8.11781 × 10⁻⁵ rad/s

The angular speed of rotation of the plane \omega_{rotation}  = 8.11781 × 10⁻⁵ rad/s

Part B

The time it takes the planet to revolve round the neighboring star once = 69.3 Earth days

Therefore, the average angular speed of the planet around its neighboring star, \omega _{Star}, is given as follows;

\omega _{Orbit}  = 2·π/((69.3 × 24 × 60 × 60) s) = 1.04938 × 10⁻⁶ rad/s

The average angular speed of orbit, \omega _{Orbit} = 1.04938 × 10⁻⁶ rad/s

The angular speed of orbit of the planet, \omega _{Orbit} = 1.04938 × 10⁻⁶ rad/s.

3 0
2 years ago
A 100 kg box is hanging from two strings. String #1 pulls up and left, making an angle of 80o with the horizontal on the left, a
frutty [35]

Answer:296.76 N

Explanation:

Given

mass of box m=100 kg

Let T_1 be the Tension in left side and T_2 be the Tension in the right side

From diagram

T_1\cos 80=T_2\cos 65

T_1=T_2\cdot \frac{\cos 65}{\cos 80}

and

T_1\sin 80+T_2\sin 65=100\cdot g

T_2\left [ \tan 80\cdot \cos 65+\sin 65\right ]=100\cdot g

T_2=\frac{100\cdot g}{\left [ \tan 80\cdot \cos 65+\sin 65\right ]}

T_2=\frac{980}{3.3023}=296.76 N

                       

3 0
3 years ago
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