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Grace [21]
2 years ago
15

A boy pushes a rock up a hill in 20 seconds. If he applied a force of 20 N over a distance of 5 m on the rock, calculate the pow

er required in this situation.
80 watts
0.2 watts
2,000 watts
5 watts
Physics
1 answer:
aksik [14]2 years ago
4 0

Here is your answer

5 watts

REASON :

Given,

t= 20 secs

F= 20 N

s (displacement)= 5m

Now,

Power= \frac{Work done}{time}

P= W/t ...... (i)

W= F×s×cos(theta)

W= 20×5 × 1 (theta is 0 degree; so cos0= 1)

W= 100 J

Hence,

P= 100/20 (from eq. i)

P= 5 watts

HOPE IT IS USEFUL

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A straight wire 20 cm long, carrying a current of 4 A, is in a uniform magnetic field of 0.6 T. What is the force on the wire wh
zheka24 [161]

Answer:

Magnetic force, F = 0.24 N

Explanation:

It is given that,

Current flowing in the wire, I = 4 A

Length of the wire, L = 20 cm = 0.2 m

Magnetic field, B = 0.6 T

Angle between force and the magnetic field, θ = 30°. The magnetic force is given by :

F=ILB\ sin\theta

F=4\ A\times 0.2\ m\times 0.6\ T\ sin(30)

F = 0.24 N

So, the force on the wire at an angle of 30° with respect to the field is 0.24 N. Hence, this is the required solution.

7 0
2 years ago
Read 2 more answers
A race car has a maximum speed of 0.104 km/s .What is this speed in miles per hour ?
sweet [91]

Answer:

232.641374 mph

Explanation:

A race car has a maximum speed of 0.104km/s

Let X represent the speed in miles per hour

Therefore the speed in miles per hour can be calculated as follows

1 km/s = 2,236.936292 mph

0.104km/s = X

X = 0.104 × 2,236.936292

X = 232.641374

Hence the speed in miles per hour is 232.641374 mph

8 0
1 year ago
A man attempts to pick up his suitcase of weight w_s by pulling straight up on the handle. (Part A figure) However, he is unable
alexira [117]

Answer:

Part A. The magnitude of the normal force is equal to the magnitude of the weight of the suitcase minus the magnitude of the force of the pull.

Part B. The magnitude of normal force acting on the suitcase is equal to the sum of the weight of the suitcase and the man.

Explanation:

Part A. This is because when the man pulls on the suit upwards, he exerts a force in the upward direction. This takes part of the force of weight of the suitcase and decreases the force the suitcase is exerting on the ground. Thus, the normal force (force exerted by suitcase on the ground) also decreases by the same force as the pull.

Part B. The statements for this part were not given in the question, but the answer reflects what is going to happen in that scenario. Since the man sits on the suitcase, the total weight acting on the ground through the suitcase is that of the suitcase plus the man. Since this force (acting on the ground) is normal force, the statement given in the answer is correct.

8 0
2 years ago
Una columna de mármol, cuya área de sección transversal es de 2.0 m2 sostiene una masa de 25.000 kg. Encontrar: (3 pto )a) El es
bazaltina [42]

Responder:

122,500 Pa; 2.45 × 10 ^ -6; 2.94 × 10 ^ -5m

Explicación:

Dado lo siguiente:

Área de sección transversal (A) = 2m ^ 2

Masa (m) = 25000 kg

Módulo de Young = 50 x 10 ^ 9 N / m2

(1) estrés en la columna:

Estrés = Fuerza / Área

F = masa * aceleración debido a la gravedad

F = 25000kg * 9.8m / s ^ 2 = 245,000J

Estrés = 245,000J / 2m ^ 2

Estrés = 122,500 Pa

2) Deformación de la unidad (deformación):

Usando la relación:

Módulo de Young = Estrés / tensión

50 × 10 ^ 9 = 122,500 / CEPA

Cepa = 122500 / (50 × 10^9)

Cepa = 0.00000245

C) Si la altura es de 12 m, ¿cuánto se acorta la columna?

Cepa = extensión / longitud

0.00000245 = extensión / 12

0,00000245 * 12

0,0000294 m

7 0
2 years ago
At one point in the rescue operation, breakdown vehicle A is exerting a force of 4000 N and breakdown vehicle B is exerting a fo
lukranit [14]

Answer:

1.) Magnitude = 5596 N

2.) Direction = 60 degrees

Explanation: You are given that the breakdown vehicle A is exerting a force of 4000 N at angle 45 degree to the vertical and breakdown vehicle B is exerting a force of 2000 N

Let us resolve the two forces into X and Y component

Sum of the forces in the X - component will be 4000 × cos 45 = 2828.43 N

Sum of the forces in the Y - component will be 2000 + ( 4000 × sin 45 )

= 2000 + 2828.43

= 4828.43 N

The resultant force R will be

R = sqrt ( X^2 + Y^2 )

Substitutes the forces at X component and Y component into the formula

R = sqrt ( 2828.43^2 + 4828.43^2 )

R = sqrt ( 31313752.53 )

R = 5595.87 N

The direction will be

Tan Ø = Y/X

Substitute Y and X into the formula

Tan Ø = 4828.43 / 2828.43

Tan Ø = 1.707106

Ø = tan^-1( 1.707106 )

Ø = 59.64 degree

Therefore, approximately, the magnitude and direction of the resultant force on the truck are 5596 N and 60 degree respectively.

8 0
2 years ago
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