What is numbers 2 and 4? Please help, and many thanks to those who do chose to help.

A man pulls a 50.0 kg box with a rope parallel to the ground he pulls the box 10.0 m about how much work has he done

anygoal [31]

Your answer is 5000 J

when W(work) = F X when F= the force and X= the displacment

and F(g) = M a(g) when M= mass and a = the acceleration and in our question
, the force is the gravitational force and a= 9.8 m/S2 we can assume as 10 m/s2

and when we have M= 50 Kg
so by substitution:
F= 50 x 10 = 500 N

and by substitution in work equation: when x = 10 m
∴ W = 500 x 10 = 5000 j

4 0

3 years ago

Read 2 more answers

75% of what number is 27? EXPLAIN

zmey [24]

75 percent (calculated percentage %) of what number equals 27? Answer: 36.

8 0

3 years ago

Read 2 more answers

Someone help me like please thank you

lianna [129]

The car should have less kinetic energy.
They are both going the same speed, but the truck is bigger and heavier. The more mass an object has, the more kinetic energy it has. There is more mass being moved, so it makes more kinetic energy. The car does not have as much mass, so it makes less kinetic energy compared to the truck.

Good luck with the rest of your test or quiz :)

3 0

3 years ago

An infinite conducting cylindrical shell of outer radius r1 = 0.10 m and inner radius r2 = 0.08 m initially carries a surface ch

irinina [24]

Answer:

a) $\sigma_{\rm in} = -2.18~{\rm \mu C/m^2}$

b) $\sigma_{\rm out}= 1.12~{\rm \mu C/m^2}$

c) $E = \frac{\sigma(r_1 + r_2)}{\epsilon_0 r}$

Explanation:

Before the wire is inserted, the total charge on the inner and outer surface of the cylindrical shell is as follows:

$Q_{\rm in} = \sigma A_{\rm in} = \sigma(2\pi r_1 h) = (-0.35)(2\pi (0.08) h) = -0.175h~{\rm \mu C}$

$Q_{\rm out} = \sigma A_{\rm out} = \sigma(2\pi r_2 h) = (-0.35)(2\pi (0.1) h) = -0.22h~{\rm \mu C}$

Here, 'h' denotes the length of the cylinder. The total charge of the cylindrical shell is -0.395h μC.

When the thin wire is inserted, the positive charge of the wire attracts the same amount of negative charge on the inner surface of the shell.

$Q_{\rm wire} = \lambda h = 1.1h~{\rm \mu C}$

a) The new charge on the inner shell is -1.1h μC. Therefore, the new surface charge density of the inner shell can be calculated as follows:

$\sigma_2 = \frac{Q_{\rm in}}{2\pi r_1h} = \frac{-1.1h}{2\pi r_1 h} = \frac{-1.1}{2\pi(0.08)} = -2.18~{\rm \mu C/m^2}$

b) The new charge on the outer shell is equal to the total charge minus the inner charge. Therefore, the new charge on the outer shell is +0.705 μC.

The new surface charge density can be calculated as follows:

$\sigma_{\rm out}= \frac{Q_{\rm out}}{2\pi r_2h} = \frac{0.705h}{2\pi r_2 h} = \frac{0.705}{2\pi(0.1)} = 1.12~{\rm \mu C/m^2}$

c) The electric field outside the cylinder can be found by Gauss' Law:

$\int{\vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}$

We will draw an imaginary cylindrical shell with radius r > r2. The integral in the left-hand side will be equal to the area of the imaginary surface multiplied by the E-field.

$E(2\pi r h) = \frac{Q_{\rm enc}}{\epsilon_0}\\E2\pi rh = \frac{\sigma 2\pi (r_1 + r_2)h}{\epsilon_0}\\E = \frac{\sigma(r_1 + r_2)}{\epsilon_0 r}$

4 0

3 years ago

A Chinook salmon can jump out of water with a speed of 6.70 m/s6.70 m/s . How far horizontally dd can a Chinook salmon travel th

aksik [14]

Answer:

R = 3.88 m

Explanation:

As the Chinook salmon leaves the water till it gets back into the water it is performing a projectile motion with the following parameters:

V₀ = Launch Speed = 6.7 m/s

θ = Launch Angle = 29°

R= Range of Projectile= Horizontal Distance Covered by Chinook salmon= ?

The value of the range of a projectile is given by the following formula:

R = (V₀² Sin 2θ)/g

R = [(6.7 m/s)² Sin {(2)(29°)}/(9.8 m/s²)]

R = [(6.7 m/s)² Sin (58°)/(9.8 m/s²)]

8 0

4 years ago