True or false forces in a force pair never cancel each other out

Differences between Pressure and upthrust

Angelina_Jolie [31]

Answer:

Pressure is equal to the ratio of thrust to the area in contact. Upthrust is a force exerted by the fluids on an object placed in the fluid . Upthrust acts in upward direction.

4 0

3 years ago

An airplane dropped a flare from a height of 2860 feet above a lake. How many seconds did it take for the flare to reach the wat

KATRIN_1 [288]

Answer: 13.2 seconds.

Explanation: using equation of motion; S= ut +1/2at² where u = initial velocity=0

S= distance travelled

a = acceleration due gravity

t= time.

1 foot = 0.305m so,

S= 2860 feet =872.3m

S= ut+1/2 at²

872.3 = 0×t + 1/2×10 × t²

872.3 =0 + 5t²

T²= 872.3/5

T²= 174.46

Take the square root of T we then have;

t = 13.2 seconds to one decimal place.

8 0

3 years ago

Read 2 more answers

What is the charge exchanged through contact ?

DIA [1.3K]

I was hoping that there would be some options to choose from in regards to this question. Since there are no options given, so i am giving the answer based on my knowledge and hope that it helps. The charge exchanged through contact is known as static charge of electrostatic charge. This charge is exchanged mostly through rubbing between two objects.

7 0

3 years ago

A kangaroo jumps straight up to a vertical height of 1.66 m. How long was it in the air before returning to Earth? Express your

Nataly [62]

Answer:

The kangaroo was 1.164s in the air before returning to Earth

Explanation:

For this we are going to use the equation of distance for an uniformly accelerated movement, that is:

$x = x_{0} + V_{0}t + \frac{1}{2}at^2$

Where:

x = Final distance

xo = Initial point

Vo = Initial velocity

a = Acceleration

t = time

We have the following values:

x = 1.66m

xo = 0m (the kangaroo starts from the floor)

Vo = 0 m/s (each jump starts from the floor and from a resting position)

a = 9.8 m/s^2 (the acceleration is the one generated by the gravity of earth)

t =This is just the time it takes to the kangaoo reach the 1.66m, we don't know the value.

Now replace the values in the equation

$x = x_{0} + V_{0}t + \frac{1}{2}at^2$

$1.66 = 0 + 0t + \frac{1}{2}9.8t^2$

$1.66 = 4.9t^2$

$\frac{1.66}{4.9} = t^2$

$\sqrt{0.339} = t\\ t = 0.582s$

It takes to the kangaroo 0.582s to go up and the same time to go down then the total time it is in the air before returning to earth is

t = 0.582s + 0.582s

t = 1.164s

The kangaroo was 1.164s in the air before returning to Earth

5 0

3 years ago

Multiply (x-3)(4x+2) using the distributive property

vazorg [7]

Are there options, because I believe the answer is D.

6 0

3 years ago