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3 years ago

7

How to calculate maximum height, when i know only V Pleeeaaassseee as fast as you can

1 answer:

Dahasolnce [82]3 years ago

8 0

Answer: was it this problem?

Explanation:

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Which is one way scientists indicate how precise and accurate there experimental measurements are

kherson [118]

They do the method 3 times to be sure. Because if you do it once, that could mean anything. If you do it twice, it may or may not have the same result. If you do it 3 times and it matches one of the previous answers, then it's likely that it's correct.

8 0

3 years ago

Two dump trucks each have a mass of 1,500 kg. The distance of the dump truck

xxTIMURxx [149]

Answer:

6.00 x 10⁻⁸N

Explanation:

Given parameters:

Mass of each dump trucks = 1500kg

Distance between them = 50m

Unknown:

New gravitational force between them = ?

Solution:

From Newton's law of universal gravitation,

F = $\frac{G m1 m2}{r^{2} }$

F is the gravitational force

G is the universal gravitation constant

m is the mass

r is the distance

F = $\frac{6.67 x 10^{-11} x 1500 x 1500}{50^{2} }$ = 6.00 x 10⁻⁸N

4 0

3 years ago

Cliff divers at Acapulco jump into the sea macias (fjm793) – Homework 3, 2d motion 19-20 – dowd – (WoffordWPHY11920 2) 4 from a

Stells [14]

Answer:

v = 7.67 m/s

Explanation:

Given data:

horizontal distance 11.98 m

Acceleration due to gravity 9.8 m/s^2

Assuming initial velocity is zero

we know that

$h = \frac{gt^2}{2}$

solving for t

we have

$t = \sqrt{\frac{2h}{g}}$

substituing all value for time t

$t = \sqrt{\frac{2\times 11.98}{9.8}}$

t = 1.56 s

we know that speed is given as

$v = \frac{d}{t}$

$v =\frac{11.98}{1.56}$

v = 7.67 m/s

7 0

3 years ago

A pulsar is a rapidly rotating neutron star. The Crab nebula pulsar in the constellation Taurus has a period of 33.5\times 10^{-

joja [24]

Answer:

$5.25\cdot 10^{40} kg m^2/s$

Explanation:

The angular momentum of the pulsar is given by:

$L=m\omega r^2$

where

$m=2.8\cdot 10^{30} kg$ is the mass of the pulsar

$r = 10.0 km = 1\cdot 10^4 m$ is the radius

$\omega$ is the angular speed

Given the period of the pulsar, $T=33.5\cdot 10^{-3} s$ , the angular speed is given by

$\omega=\frac{2\pi}{T}=\frac{2 \pi}{33.5\cdot 10^{-3}s}=187.5 rad/s$

And so, the angular momentum is

$L=m\omega r^2=(2.8\cdot 10^{30}kg)(187.5 rad/s)(1\cdot 10^4 m)^2=5.25\cdot 10^{40} kg m^2/s$

8 0

3 years ago

Determine the magnitude of the electrostatic force on a 0.06000 C charged object when it is placed in an electric field of magni

joja [24]

Answer:

Explanation:

Use the following equation:

$E=\frac{q}{F}$ and solve for F:

$F=\frac{q}{E}$ and filling in:

$F=\frac{.0600}{1500}$

F = 4.0 × $10^{-4$ N

3 0

3 years ago