The acceleration of gravity on the

A river flows due south with a speed of 2.0 m/s .You steer a motorboat across the river; your velocity relative to the water is

mihalych1998 [28]

Answer:

a) $v_m =\sqrt{v^2_x + v^2_y} = \sqrt{(2m/s)^2 +(4.8 m/s)^2}= 5.2 m/s$

b) $\theta= tan^{-1} \frac{v_y}{v_x} = tan^{-1} (\frac{2}{4.8})= 22.62$ degrees and on this case to the South of the East.

c) $t= \frac{w}{v_m}= \frac{600m}{4.8 m/s}= 125 s$

d) $Y = 2 m/s * 125 s = 250m$

So it would be 250 to the South

Explanation:

Part a

For this case the figure attached shows the illustration for the problem.

We know that $v_y = 2 m/s$ represent the velocity of the river to the south.

We have the velocity of the motorboard relative to the water and on this case is $V_x= 4.8 m/s$

And we want to find the velocity of the motord board relative to the Earth $v_m$

And we can find this velocity from the Pythagorean Theorem.

$v_m =\sqrt{v^2_x + v^2_y} = \sqrt{(2m/s)^2 +(4.8 m/s)^2}= 5.2 m/s$

Part b

We can find the direction with the following formula:

$\theta= tan^{-1} \frac{v_y}{v_x} = tan^{-1} (\frac{2}{4.8})= 22.62$ degrees and on this case to the South of the East.

Part c

For this case we can use the following definition

$D = Vt$

The distance would be D = w = 600 m and the velocity V = 4.8m/s and if we solve for t we got:

$t= \frac{w}{v_m}= \frac{600m}{4.8 m/s}= 125 s$

Part d

For this case we can use the same definition but now using the y compnent we have:

$Y = v_y t$

And replacing we got:

$Y = 2 m/s * 125 s = 250m$

So it would be 250 to the South

7 0

3 years ago

3) What is the weight of an object with a mass of 5.2 kilograms?

inna [77]

Answer:

50.96 N

Explanation:

weight = mass x gravity

We know that gravity = 9.8 m/s^2 and mass = 5.2 kg.

w = m x g

w = 5.2 kg x 9.8 m/s^2

w = 50.96 N

The weight of the object is 50.96 N (newtons). Hope this helps, thank you !!

8 0

3 years ago

What is the measure of the force that gravity applies to an object

GarryVolchara [31]

Answer:

Gravity, or gravitation, is a natural phenomenon by which all things with mass or energy-including planets,stars,galaxies, and even light-are brought toward one another.

Explanation:

4 0

3 years ago

A proton and an electron in a hydrogen atom are separated on the average by about 5.3 × 10−11 m. What is the magnitude and direc

Genrish500 [490]

Answer:

1. 5.12068 × 1011 N/C away from the proton

Explanation:

The electric field produced by a single point charge is given by:

$E=k\frac{q}{r^2}$

where

k is the Coulomb's constant

q is the magnitude of the charge

r is the distance from the charge

In this problem, we have:

$q=1.6\cdot 10^{-19}C$ is the charge of the proton

$r=5.3\cdot 10^{-11} m$ is the distance at which we want to calculate the field

$k=8.99\cdot 10^9 Nm^2C^{-2}$ is the Coulomb's constant

Substituting into the formula,

$E=(8.99\cdot 10^9 Nm^2C^{-2})\frac{1.6\cdot 10^{-19}C}{(5.3\cdot 10^{-11}m)^2}=5.12068\cdot 10^{11} N/C$

And the direction of the electric field produced by a positive charge is away from the charge, so the correct answer is

1. 5.12068 × 1011 N/C away from the proton

4 0

4 years ago

Spiderman, whose mass is 74.0 kg, is dangling on the free end of a 11.0-m-long rope, the other end of which is fixed to a tree l

Anestetic [448]

Answer:

W = -1844.513 J

Explanation:

GIVEN DATA:

mass of spider man is m 74 kg

vertical displacement if spider is 11 m

final displacement = 11 cos 60.6 = - 6.753 m

change in displacement is = -6.753 - (-11) = 4.25 m

gravity force act on spiderman is f = mg = 74 × 9.8 = 725.2 N

work done by gravity is $W = F \delta r cos\theta$

$W = 725.2 \times 4.25 \times cos 180$

where 180 is the angle between spiderman weight and displacement

W = -1844.513 J

7 0

3 years ago