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2 years ago

A mailman performed 296 J of work lifting a box of 1.6 m. How much force did the mailman use?

Physics

1 answer:

Rus_ich [418]2 years ago

7 0

185 newtons because (W=Fd so W/d=F)296/1.6=185

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A 2.0-kg block slides on a rough horizontal surface. A force (magnitude P = 4.0 N) acting parallel to the surface is applied to

irinina [24]

When the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².

<h3>Frictional force between the block and the horizontal surface</h3>

The frictional force between the block and the horizontal surface is determined by applying Newton's law;

∑F = ma

F - Ff = ma

Ff = F - ma

Ff = 4 - 2(1.2)

Ff = 4 - 2.4

Ff = 1.6 N

When the applied force increases to 5 N, the magnitude of the block's acceleration is calculated as follows;

F - Ff = ma

5 - 1.6 = 2a

3.4 = 2a

a = 3.4/2

a = 1.7 m/s²

Thus, when the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².

Learn more about frictional force here: brainly.com/question/4618599

8 0

1 year ago

How many cups of water should you drink in a day?

Harrizon [31]

Answer:

about 2.7liters for women and 3.7liters for men

Explanation:

6 0

2 years ago

What will happen if our atmosphere was oxygen

qaws [65]

Most humans would die do to the excess oxyegen

3 0

2 years ago

A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication

koban [17]

Answer:

r = 4.24x10⁴ km.

Explanation:

To find the radius of such an orbit we need to use Kepler's third law:

$\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}$

<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km. </em>

From equation (1), r₁ is:

$r_{1} = r_{2} \sqrt[3] {(\frac{T_{1}}{T_{2}})^{2}}$

$r_{1} = 3.84\cdot 10^{5} km \sqrt[3] {(\frac{1 d}{0.07481 y \cdot \frac{365 d}{1 y}})^{2}}$

$r_{1} = 4.24 \cdot 10^{4} km$

Therefore, the radius of such an orbit is 4.24x10⁴ km.

I hope it helps you!

3 0

3 years ago

Consider a venturi with a small hole drilled in the side of the throat. This hole is connected via a tube to a closed reservoir.

Marina CMI [18]

Answer:

( $P_1-P_2$ )=1913.31 N/m^2

Explanation:

given:

$\frac{A_t}{A_1}$ =0.85

$V_1$ =90 m/s

γ∞=1.23 kg/m^3

solution:

since outside pressure is atm pressure vaccum can be defined by ( $P_1-P_2$ )

$V_1$ =√2( $P_1-P_2$ )/γ∞[ $\frac{A_t}{A_1}^2$ -1]

( $P_1-P_2$ )=1913.31 N/m^2

6 0

2 years ago