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4 years ago

A force of 20N acts on a 3.5kg mass for 10s. What is the change in the speed of the object? Hint find the impulse first

Physics

1 answer:

Elina [12.6K]4 years ago

6 0

Answer:

<h2>I don't know sorry bro .. .... please make me braniest </h2>

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Chapter 21, Problem 43Partially correct answer. Your answer is partially correct. Try again.The 1400-turn coil in a dc motor has

miskamm [114]

Answer:

I = 3.028 A

Explanation:

given,

Number of turns in the coil = N = 1400

Area of DC motor per turn = 1.1 × 10⁻² m²

Maximum torque in motor = 8.4 N.m

Magnetic field in the coil = 0.18 T

Current in the coil = ?

using formula for toque in the magnetic field

τ = N A B I

8.4 = 1400 x 1.1 × 10⁻² x 0.18 x I

$I = \dfrac{8.4}{2.774}$

I = 3.028 A

Current in the coil will be equal to I = 3.028 A

7 0

3 years ago

Does anyone know the answer

scoundrel [369]

No put false I am right

7 0

3 years ago

While unrealistic, we will examine the forces on a leg when one falls from a height by approximating the leg as a uniform cylind

Leno4ka [110]

Answer:

Part A: 7.75 m/s

Part B: 2330.8 kN

Part C: 24.03 kN

Part D: 4.8 kN

Part E: $1.7\times 10^{9} Dyn/cm^{2}$

Part F: Option D

Bending his legs increases the time over which the ground applies force, thus decreasing the force applied by the ground.

Explanation:

Part A

From the fundamental kinematic equation

$v^{2}=u^{2}+2gh$ where v is the velocity of the man just before hitting the ground, g is acceleration due to gravity, u is initial velocity, h is the height.

Since the initial velocity is zero hence

$v^{2}=2gh$

$v=\sqrt 2gh$

Substituting 10 m/s2 for g and 3 m for h we obtain

$v=\sqrt 2\times 10\times 3 =\sqrt 60= 7.745967\approx 7.75 m/s$

Part B 

Force exerted by the leg is given by

F=PA where P is pressure, F is force, A is the cross-section of the bone

$A=\frac {\pi d^{2}}{4}$

Substituting 2.3 cm which is equivalent to 0.023m for d and $1.7\times10^{8} N/m2$ for P we obtain the force as

$F=PA=1.7\times10^{8}*\frac {\pi (0.023)^{2}}{4}= 2330818.276\approx 2330.8 kN$

Part C

The fundamental kinematic equation is part (a) can also be written as

$v^{2}=u^{2}+2a\triangle x$ and making a the subject then

$a=\frac {v^{2}-u^{2}}{2\triangle x}$ where a is acceleration and $\triangle x$ is the change in length

Substituting the value obtained in part a, 7.75 m/s for v, u is zero and 1cm which is equivalent to 0.01 m for $\triangle x$ then

$a=\frac {7.75^{2}-0^{2}}{2\times 0.01}= 3003.125 m/s^{2}$

Force exerted on the man is given by

$F=ma=80\times 3003.125= 240250 N\approx 24.03 kN$

Part D

The fundamental kinematic equation is part (a) can also be written as

$v^{2}=u^{2}+2a\triangle h$ and making a the subject then

$a=\frac {v^{2}-u^{2}}{2\triangle h}$ where a is acceleration and $\triangle h$ is the change in height

Also, force exerted on the man is given by $F=ma=m\times \frac {v^{2}-u^{2}}{2\triangle h}$

Substituting 80 Kg for m, 50 cm which is equivalent to 0.5m for \triangle h and other values as used in part c

$F=ma=m\times \frac {v^{2}-u^{2}}{2\triangle h}=80\times \frac {7.75^{2}-0^{2}}{2\times 0.5}= 4805 N\approx 4.8 kN$

Part E

$P=1.7\times 10^{8}=1.7\times 10^{8}\times (\frac {10^{5} Dyn}{10^{4} cm^{2}}=1.7\times 10^{9} Dyn/cm^{2}$

Part F

Bending his legs increases the time over which the ground applies force, thus decreasing the force applied by the ground

7 0

3 years ago

Find the acceleration of an 800-kg car that has a net force of 4,000 N acting upon it.

kotykmax [81]

Answer:

The acceleration of the car is 5m/s²

Explanation:

Given data

Force F= 4,000N

Mass of car m= 800kg

Acceleration a= ?

Applying the Newton's second law of motion which states that the rate of change of momentum of a body is proportional to the external force and takes place in the direction of the force

F=ma

a= F/m

a= 4000/800

a=5m/s²

3 0

4 years ago

A 56 kg astronaut stands on a bathroom scale inside a rotating circular space station. The radius of the space station is 250 m.

Zielflug [23.3K]

Answer:

The speed of space station floor is 49.49 m/s.

Explanation:

Given that,