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andrezito [222]
3 years ago
11

How much tension must a rope withstand if it is used to accelerate a 1210- kg car horizontally along a frictionless surface at 1

.20 m/s squared?
Physics
1 answer:
Ratling [72]3 years ago
4 0
Newton's second law states that the force F applied to the car is the product between the mass of the car m and its acceleration a:
F=ma
But the force applied to the car is the tension of the rope, T, so we have:
T=ma

And so, using the data of the problem we calculate the value of the tension:
T=ma=(1210 kg)(1.20 m/s^2)=1452 N
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Answer:

0.16joules

Explanation:

Using the relation for The gravitational potential energy

E= Mgh

Where,

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h = Vertical Height

M = mass

g = Gravitational Field Strength

To find the vertical component of angle of launch Where the angle is 22°

h= sin theta

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= 0.18 x 0.98 x 0.253 sin22

=0.16joules

Explanation:

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Which statement best defines constructive interference? O Energy reflects back toward the source of its power. O Two waves with
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Explanation:

My teacher gave me the answer

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A solid sphere has a radius of 0.200 m and a mass of 150.0 kg. how much work is required to get the sphere rolling with an angul
Allisa [31]

Here in this case we can use work energy theorem

As per work energy theorem

Work done by all forces = Change in kinetic Energy of the object

Total kinetic energy of the solid sphere is ZERO initially as it is given at rest.

Final total kinetic energy is sum of rotational kinetic energy and translational kinetic energy

KE = \frac{1}{2}Iw^2 +\frac{1}{2} mv^2

also we know that

I = \frac{2}{5}mR^2

w= \frac{v}{R}

Now kinetic energy is given by

KE = \frac{1}{2}(\frac{2}{5}mR^2)w^2 +\frac{1}{2} m(Rw)^2

KE = \frac{1}{5}mR^2w^2 +\frac{1}{2} mR^2w^2

KE = \frac{7}{10}mR^2w^2

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KE = 10500 J

Now by work energy theorem

Work done = 10500 - 0 = 10500 J

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3 years ago
The greatest speed recorded by a baseball thrown by a pitcher was 162.3 km / h, obtained by Nolan Ryan in 1974. If the ball leav
Pepsi [2]

Answer:

0.96 m

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Δx = v₀ t + ½ at²

20 m = (45.08 m/s) t + ½ (0 m/s²) t²

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Finally, find how far the ball falls in that time.

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Δy = -0.96 m

The ball will have fallen 0.96 meters.

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4 years ago
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