Help me with this please

With your hand parallel to the floor and your palm upright, you lower a 3-kg book downward. If the force exerted on the book by

lara31 [8.8K]

According to the net force, the acceleration of the book is 16.47 m/s².

We need to know about force to solve this problem. According to second Newton's Law, the force applied to an object will be proportional to mass and acceleration. Hence, it can be written as

∑F = m . a

where F is force, m is mass and a is acceleration

From the question above, we know that

m = 3 kg

g = 9.8 m/s²

F1 = 20 N

Find the net force

∑F = F1 + W

∑F = 20 + m . g

∑F = 20 + 3 . 9.8

∑F = 20 + 29.4

∑F = 49.4 N

Find the acceleration

∑F = m . a

49.4 = 3 . a

a = 16.47 m/s²

Find more on force at: brainly.com/question/25239010

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7 0

1 year ago

A person's body is producing energy internally due to metabolic processes. If the body loses more energy than metabolic processe

SVETLANKA909090 [29]

Answer:

$T_s = 6.8 degree C$

Explanation:

As per thermal radiation we know that rate is heat radiation is given as

$\frac{dQ}{dt} = \sigma eA (T^4 - T_s^4)$

here we know that

T = 34 degree C = 307 K

$A = 1.38 m^2$

e = 0.557

$\sigma = 5.67 \times 10^{-8} W/m^2K^4$

$\frac{dQ}{dt} = 120 J/s$

now we have

$120 = (5.67 \times 10^{-8})(0.557)(1.38)(307^4 - T_s^4)$

$120 = (4.36 \times 10^{-8})(307^4 - T_s^4)$

$T_s = 279.8 K$

$T_s = 6.8 degree C$

5 0

3 years ago

A cart traveling at 0.3 m/s collides with stationary object. After the collision, the cart rebounds in the opposite direction. T

miv72 [106K]

Answer:

C. In the first collision has twice the momentum as when it stays still ( second colllions)

Explanation:

To see which statement is correct, it is best to solve the problem, the momentum is equal to the variation of the moment

I = Δp = m vf - m v₀

I = m (vf -v₀)

Case 1. In car bounces, the initial speed is 0.3 m / s, say that this direction is positive, when the magnitude of the speed bounces it remains constant, but its direction is reversed (vf = -0.3 m / s)

I₁ = m (-0.3 - 0.3)

I₁ = -0.6 m

Case 2. The expensive one that still after the crash so its speed is zero (vf = 0)

I₂ = m (0 - 0.3)

I₂ = -0.3 m

Let's calculate the relationship between the two impulses

I₁ / I₂ = -0.6m / -0.3m

I₂ / I₂ = 2

When it bounces it has twice the momentum as when it stays still

Now let's analyze the answers:

A. False The momentum changes

B. False. The momentum is less in the second collision

C. True. The momentum is double in this collision

D. False. Can be calculated, because the mass is the same throughout the exercise and is eliminated in the equations

E. False. When they say bounces it implies the same speed with the opposite direction

4 0

3 years ago

(20 points) Pressure is a measure of force per unit area or P=F/A. A student is checking the amount of pressure their car exerts

Gala2k [10]

Answer:

35.2 ps

Explanation:

Hello!

A US short ton is equal to 2 000 pounds, therefore:

2.25 tons = 4500 pounds

Now lets convert the area of each tire footprint:

1 cm = 0.393701 in

12.5 cm = 4.92126 in

16.5 cm = 6.49607 in

Therefore, the area of the footprint is: 31.9688 in^2

Then, the pressure exerted by the car is given by the weigth devided by the total area, that is 4 times the area of the footprint :

4500 /(4 * 31.9688) psi = 35.1906 psi

Rounding to the first tenth

35.2 psi

6 0

3 years ago

19

PtichkaEL [24]

Answer: The mass of ball is 10 grams.

Explanation:

Given : Kinetic energy = 4.5 J

Velocity = 30 m/s

The formula for Kinetic energy is as follows.

$K.E = \frac{1}{2}mv^{2}$

where,

K.E = Kinetic energy

m = mass

v = velocity

Substitute the values into above formula as follows.

$K.E = \frac{1}{2}mv^{2}\\4.5 J = \frac{1}{2} \times m \times (30 m/s)^{2}\\m = \frac{4.5 J \times 2}{900 m^2/s^2} (1 Js^{2} = kg m^{2})\\= 0.01 kg (1 kg = 1000 g)\\= 10 g$

Thus, we can conclude that the mass of ball is 10 grams.

3 0

3 years ago