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2 years ago

8

What is the equation to find an angle of projectile

1 answer:

Ksju [112]2 years ago

3 0

I do not recall the answer to this question

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Assuming 70% of Earth's surface is covered in water at an average depth of 2.5 mi, estimate the mass of the water on Earth in Ki

saw5 [17]

This is an excellent question that i do not have the answer to.

7 0

2 years ago

What is a moment of a force

kirza4 [7]

Answer:

In physics and mechanics, torque is the rotational equivalent of linear force. It is also referred to as the moment, moment of force, rotational force or turning effect, depending on the field of study. The concept originated with the studies by Archimedes of the usage of levers

7 0

2 years ago

What is the gravitational force on a 70kg that is 6.38x10^6m above the earths surface

NARA [144]

Answer:

171.5 N

Explanation:

The gravitational force on an object due to the Earth is given by

$F=mg$

where

m is the mass of the object

g is the acceleration due to gravity

The acceleration due to gravity at a certain height h above the Earth is given by

$g=\frac{GM}{(R+h)^2}$

where:

G is the gravitational constant

$M=5.98\cdot 10^{24} kg$ is the Earth's mass

$R=6.37\cdot 10^6 m$ is the Earth's radius

Here,

$h=6.38\cdot 10^6 m$

So the acceleration due to gravity is

$g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})}{(6.37\cdot 10^6 + 6.38\cdot 10^6)^2}=2.45 m/s^2$

We know that the mass of the object is

m = 70 kg

So, the gravitational force on it is

$F=mg=(70)(2.45)=171.5 N$

5 0

3 years ago

Look at the graph. What is the slope of the line?

NeTakaya

For the first hour, the slope is zero.

After that, the slope is -2 miles per hour.

7 0

2 years ago

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

neonofarm [45]

Answer:

a) 25.5°(south of east)

b) 119 s

c) 238 m

Explanation:

solution:

we have river speed $v_{r}$ =2 m/s

velocity of motorboat relative to water is $v_{m/r}$ =4.2 m/s

so speed will be:

a) $v_{m}$ = $v_{r}$ + $v_{m/r}$

solving graphically

$v_{m}=\sqrt{v^2_{r}+v^2_{m/r}}$

=4.7 m/s

Ф= $tan^{-1} (\frac{v_{r}}{v_{m/r}} )$

=25.5°(south of east)

b) time to cross the river: t= $\frac{w}{v_{m/r}}$ = $\frac{500}{4.2}$ =119 s

c) d= $v_{r}t$ =(2)(119)=238 m

note :

pic is attached

6 0

3 years ago