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3 years ago

Paco pulls a 67 kg crate with 738 N and of force across a frictionless floor 9.0 M how much work does he do in moving the crate

Physics

1 answer:

olga_2 [115]3 years ago

5 0

Answer:

W = 6642 J

Explanation:

Given that,

Mass of a crate, m = 67 kg

Force with which the crate is pulled, F = 738 N

It is moved 9 m across a frictionless floor

We need to find the work done in moving the crate. Let the work done is W. It is given by :

W = F d

W = 738 N × 9 m

= 6642 J

So, the work done is 6642 J.

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Find the work done in pumping gasoline that weighs 6600 newtons per cubic meter. A cylindrical gasoline tank 3 meters in diamete

Sonbull [250]

Answer:

work done in pumping the entire fuel is 466587 J

Explanation:

weight of the gasoline per volume = 6600 N/m^3

diameter of the tank = 3 m

length of the tank = 2 m

height of the tractor tank above the top of the tank = 5 m

work done in pumping fuel to this height = ?

First, we find the volume of the fuel

since the tank is cylindrical, we assume that the fuel within also takes the cylindrical shape.

Also, we assume that the fuel completely fills the tank.

volume of a cylinder = $\pi r^{2}l$

where r = radius = diameter ÷ 2 = 3/2 = 1.5 m

volume of the cylinder = 3.142 x $1.5^{2}$ x 2 = 14.139 m^3

we then find the total weight of the fuel in Newton

total weight = (weight per volume) x volume

total weight = 6600 x 14.139 = 93317.4 N

work done = (total weight of the fuel) x (height through which the fuel is pumped)

work done in pumping = 93317.4 x 5 = 466587 J

8 0

3 years ago

Determine the momentum of a 15,000 kg truck traveling at 15 m/s

S_A_V [24]

Answer:

mass*velocity=1.5*10^4 * 15

= 22.5*10^4

6 0

3 years ago

a length of wire is cut into five equal pieces. the five pieces are then connected in parallel, with the resulting resistance be

Natali [406]

The equivalent resistance of n resistors connected in parallel is given by
$\frac{1}{R_{eq}} = \frac{1}{R_1}+ \frac{1}{R_2}+...+ \frac{1}{R_n}$ (1)

In our problem, the resulting resistance of the 5 pieces connected in parallel is $R_{Eq}=2.00 \Omega$ , and since the 5 pieces are identical, their resistance R is identical, so we can rewrite (1) as
$\frac{1}{R_{Eq} }= \frac{1}{2 \Omega}= \frac{1}{R}+ \frac{1}{R} + \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{5}{R}$
From which we find $R= 5 \cdot 2 \Omega = 10 \Omega$ .

So, each piece of wire has a resistance of $10 \Omega$ . Before the wire was cut, the five pieces were connected as they were in series. The equivalent resistance of a series of n resistors is given by
$R_{Eq}=R_1 + R_2 + ...+R_n$
So if we apply it at our case, we have
$R_{eq}=R+R+R+R+R=5 R= 5\cdot 10 \Omega= 50 \Omega$

therefore, the resistance of the original wire was $50 \Omega$ .

5 0

3 years ago

A car travels at a speed of 55 km/hr and slows down to 10 km/hr in 20 seconds. What is the acceleration?

WITCHER [35]

Answer:

Explanation:

3 0

3 years ago

Read 2 more answers

An egg is thrown nearly vertically upward from a point near the cornice of a tall building. The egg just misses the cornice on t

diamong [38]

Answer:

1. 18.5m/s

2. 17.5 m

3. 0 at its highest point

4. Direction is downwards

Explanation:

1. This egg is thrown vertically from a height

Yo = 0. This egg then falls to the point y = -30.0 at t = 5seconds

Y-Yo = V0t - 1/2gt²

-30-0 = V0(5)-1/2(9.8)(5²)

-30 = 5v0 - 4.9x25

-30 = 5V0 - 122.5

-30+122.4 = 5v0

V0 = 92.5/5

= 18.5m/s

this is the initial speed of the egg

2. When the egg is at a maximum height it would have a velocity equal to 0

V² = V0² - 2*g*y

V = 0, V0 = 18.5, g = 9.8

0 = 18.5²-2x9.8*y

342.25-19.6y = 0

342.25 = 19.6y

Divide through by 19.6

Y = 342.25/19.6

Y = 17.5m

this value is how high it rises above starting point

The magnitude of velocity is = 0 at its highest point

This egg falls under gravity. Therefore the acceleration due to gravity has a constant magnitude and direction. Magnitude = 9.8m/s and it's direction is downwards.

5. Please check attachment for graph

6 0

3 years ago