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olganol [36]
2 years ago
11

If a 1.50 kg mass revolves at the end of a string 0.50 m long, and its tangential speed is 6.0 m/s, calculate the centripetal fo

rce.
Physics
2 answers:
Allisa [31]2 years ago
8 0

Answer:

108\ N

Explanation:

Mass of object, m=1.50\ kg.

Length of string, r=0.50\ m.

Tangential speed, v_t=6.0\ m/s.

Now, centripetal force F of a object moving in given radius r and mass m moving with velocity v.

Here , object moves along the ends of string. Therefore, radius is equal to length of string.

F=\dfrac{m \times v_t^2}{r}.

Putting values of m,v and r in above equation.

We get, F=\dfrac{1.50\times (6.0)^2}{0.50} \ N=108 \ N.

Hence, this is the required solution.

rosijanka [135]2 years ago
5 0
You should get about 110 for an answer
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Um navio cargueiro proveniente do Oceano Atlântico passa a navegar nas águas menos densas do rio Amazonas. Em comparação com a s
LUCKY_DIMON [66]

Answer:

Mas eu acho q é o empuxo será igual e a porção imersa do navio será maior.

Explanation: SE

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8 0
2 years ago
A 56 kg diver runs and dives from the edge of a cliff into the water which is located 4.0 m below. If she is moving at 8.0 m/s t
Reil [10]

Answer:

1) 2197.44 J

2) 0 J

3) 2197.44 J = Constant

4) 2197.44 J

5) Approximately 8.86 m/s

Explanation:

The given parameters are;

The mass of the diver, m = 56 kg

The height of the cliff, h = 4.0 m

The speed with which the diver is moving, vₓ = 8.0 m/s

The gravitational potential energy = Mass, m × Height of the cliff, h × Acceleration due to gravity, g

1) Her gravitational potential energy = 56 × 4.0 × 9.81 = 2197.44 J

2) The kinetic energy = 1/2·m·u²

Where;

u = Her initial velocity = 0 when she just leaves the cliff

Therefore;

Her kinetic energy when she just leaves the cliff = 1/2 × 56 × 0² = 0 J

3) The total mechanical energy = Kinetic energy + Potential energy

The total mechanical energy is constant

Her total mechanical energy relative to the water surface when she leaves the cliff = Her gravitational potential energy = 2197.44 J = Constant

4) Her total mechanical energy relative to the water surface just before she enters the water = 2197.44 J

5) The speed with which she enters the water, v, is given from, v² = u² + 2·g·h

Where;

u = The initial velocity at the top of the cliff before she jumps= 0 m/s

∴ v² = 0² + 2 × 9.81 × 4 = 78.48

v = √78.48 ≈ 8.86 m/s

The speed with which she enters the water, v ≈ 8.86 m/s

7 0
2 years ago
A student produces a power of p = 0.87 kw while pushing a block of mass m = 75 kg on an inclined surface making an angle of θ =
Paul [167]

When block is pushed upwards along the inclined plane

the net force applied on the block will be given as

F_{net} = mg sin\theta + \mu_k mg cos\theta

here we know that

m = 75 kg

\theta = 8.5 degree

\mu_k = 0.16

now plug in all values into this

F_{net} = 75\times 9.8 sin8.5 + 0.16 \times 75\times 9.8 cos8.5

F = 225 N

now for finding the power is given as

P = Fv

0.87 \times 10^3 = 225 \time v

v = \frac{870}{225} = 3.87 m/s

6 0
2 years ago
A car has a mass of 1600 kg. It is stuck in the snow and is being pulled out by a cable that applies a force of 7560 N due north
sergiy2304 [10]

Answer:

Acceleration of the car will be a=0.1375m/sec^2

Explanation:

We have given mass of the ball m = 1600 kg

Force in north direction F= 7560 N

Resistance force which opposes the movement of car F_R=7340N

So net force on the car F_{net}=F-F_R=7560-7340=220N

According to second law of motion we know that F=ma

So 220=1600\times a

a=0.1375m/sec^2

7 0
2 years ago
An object undergoing simple harmonic motion has a maximum displacement of 6.2 m at t=0.0 s. if the angular frequency of oscillat
Law Incorporation [45]

Answer:

C

Explanation:

From above question we know that

A = 6.2 m

f = 1.6 rad/s

t = 3.5 s

x =?

We know that,

x = Acos(2pie ft)

Putting all values in above eq.

x = 6.2 x cos(2x3.142x1.6x3.5)

x = - 4.8

Displacement can never be negative so ignore - sign.

4 0
2 years ago
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