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bearhunter [10]
2 years ago
15

A construction worker pushes a crate horizontally on a frictionless floor with a net force of 10\, \text 10N, start text, N, end

text for 4.0\,\text m4. 0m4, point, 0, start text, m, end text. How much kinetic energy does the crate gain?
Physics
1 answer:
Strike441 [17]2 years ago
7 0

Answer:

k_f = 40J

Explanation:

The work and energy theorem says that:

W_f =k_f-k_i

where W_f is the work of the force, k_f the final kinetic energy and k_i the initial kinetic energy.

Addittionally, the work of the force is calculate as force multiply by distance and if the crate inittialy is at rest, the initial kinetic energy is zero, so:

Fd = k_f

where F is the force and d the distance. Then, replacing values, we get:

(10N)(4m) = k_f

40J = k_f

it means that the system gain 40J of kinetic energy.

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Imagine you derive the following expression by analyzing the physics of a particular system: v2=v20+2ax. The problem requires so
lisabon 2012 [21]

As per kinematics equation we are given that

v^2 = v_o^2 + 2ax

now we are given that

a = 2.55 m/s^2

v_0 = 21.8 m/s

v = 0

now we need to find x

from above equation we have

0^2 = 21.8^2 + 2(2.55)x

0 = 475.24 + 5.1 x

x = 93.2 m

so it will cover a distance of 93.2 m

7 0
2 years ago
Read 2 more answers
Two flywheels of negligible mass and different radii are bonded together and rotate about a common axis (see below). The smaller
jeka94

Answer:

Explanation:

Torque on smaller wheel

= F x r

50 x .30

= 15 Nm

Torque on larger wheel

= F x .5

For equilibrium

F x .5 = 15

F = 15 / .5

= 30 N

8 0
2 years ago
In general it is best to conceptualize vectors as arrows in space, and then to make calculations with them using their component
Dimas [21]

(1) A - B

(2) B - C

(3) - A + B - C

(4) 3A - 2C

(5) - 2A + 3B - C

(6) 2A - 3 (B - C)

Answer:

(1)  (3,-5,-4)

(2) (-5, 4, 0)

(3) (-6, 4, 3)

(4) (-3, -2, -11)

(5) (-11, 14, 8)

(6) (17, -12, -6)

Explanation:

A⃗ =(1,0,−3)

B⃗ =(−2,5,1)

C⃗ =(3,1,1)

Vector additions and subtraction are done on a component by component basis, that is, only data from component î can be added to or subtracted from another Vector's component î. And so on for components j and k.

1) (A - B) = (1,0,−3) - (−2,5,1) = (1-(-2), 0-5, -3-1) = (3,-5,-4)

2)  (B - C) = (−2,5,1) - (3,1,1) = (-2-3, 5-1, 1-1) = (-5, 4, 0)

3) -A + B - C = -(1,0,−3) + (−2,5,1) - (3,1,1) = (-1-2-3, 0+5-1, 3+1-1) = (-6, 4, 3)

4) 3A - 2C = 3(1,0,−3) - 2(3,1,1) = (3,0,-9) - (6,2,2) = (3-6, 0-2, -9-2) = (-3, -2, -11)

5) -2A + 3B - C = -2(1,0,−3) + 3(−2,5,1) - (3,1,1) = (-2,0,6) + (-6,15,3) - (3,1,1) = (-2-6-3, 0+15-1, 6+3-1) = (-11, 14, 8)

6) 2A - 3 (B - C) = 2(1,0,−3) - 3[(−2,5,1) - (3,1,1)] = (2,0,-6) - 3(-5,4,0) = (2+15, 0-12, -6-0) = (17, -12, -6)

3 0
2 years ago
A truck collides with a car on horizontal ground. At one moment during the collision, the magnitude of the acceleration of the t
Mice21 [21]

Answer:

The magnitude of the acceleration of the car is 35.53 m/s²

Explanation:

Given;

acceleration of the truck, a_t = 12.7 m/s²

mass of the truck, m_t = 2490 kg

mass of the car, m_c = 890 kg

let the acceleration of the car at the moment they collided = a_c

Apply Newton's third law of motion;

Magnitude of force exerted by the truck = Magnitude of force exerted by the car.

The force exerted by the car occurs in the opposite direction.

F_c = -F_t\\\\m_ca_c = -m_t a_t\\\\a_c =- \frac{m_ta_t}{m_c} \\\\a_c = -\frac{2490 \times 12.7}{890} \\\\a_c = - 35.53 \ m/s^2

Therefore, the magnitude of the acceleration of the car is 35.53 m/s²

3 0
2 years ago
You decide to work at a heart rate of 150 instead of 120. What area of F.I.T.T. did you change?
Rina8888 [55]

Key concepts

Heart rate

Exercising

The heart

Cardiovascular system

Health

Introduction

As Valentine's Day approaches, we're increasingly confronted with "artistic" images of the heart. Real hearts hardly resemble to two-lobed shapes adorning cards and candy boxes this time of year. And the actual shape of the human heart is important for its function of supplying blood to the entire body. You have likely noticed that your heart beats more quickly when you exercise. But have you ever taken the time to observe how long it takes to return to its normal rate after you're done exercising? In this science activity you'll get to do some exercises to explore your own heart-rate recovery time.

Background

Your heart is continuously beating to keep blood circulating throughout your body. Its rate changes depending on your activity level; it is lower while you are asleep and at rest and higher while you exercise—to supply your muscles with enough freshly oxygenated blood to keep the functioning at a high level. Because your heart is also a muscle, exercise, in turn, helps keep it healthy. The American Heart Association recommends that a person does exercise that is vigorous enough to raise their heart rate to their target heart-rate zone—50 percent to 85 percent of their maximum heart rate, which is 220 beats per minute (bpm) minus their age for adults—for at least 30 minutes on most days, or about 150 minutes a week in total. So for a 20-year-old, the maximum heart rate would be 200 bpm, with a target heart-rate zone of 100 to 170 bpm. (For those 19 or younger, target zones can vary more than they do for adults.)

i think it will help you...if it help you ...please mark brainless

8 0
2 years ago
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