The magnitude of a force is:

What is the approximate weight of a 20-kg cannonball on earth?

disa [49]

The weight of an object is taken to be the force on the object due to gravity. The weight ( W ) is the product of the mass ( m ) of the object and the magnitude of the gravitational acceleration ( g ).
On Earth: g = 9.81 m/s²
m = 20 kg
W = m · g = 20 kg · 9.81 m/s² = 196.2 N

6 0

3 years ago

You drop a rock from rest out of a window on the top floor of a building, 20.0 m above the ground. When the rock has fallen 5.00

Andrej [43]

Answer:

You drop a rock from rest out of a window on the top floor of a building, 30.0 m above the ground. When the rock has fallen 3.00 m, your friend throws a second rock straight down from the same window. You notice that both rocks reach the ground at the exact same time. What was the initial velocity of the ...... rest out of a window on the top floor of a building, 30.0m above the ground. ... You Notice That Both Rocks Reach The Ground At The Exact Same Time. ... You drop a rock from rest out of a window on the top floor of a building, 30.0m ... When the rock has fallen 3.20 m, your friend throws a second rock straight down from ...

4 0

2 years ago

an object is moving with initial velocity of 5 m/s. After 10 seconds final velocity is 10 m/s. Calculate its acceleration.

Brut [27]

Answer:

0.5m/s2

Explanation:

acceleration= change in velocity/time taken

= v - u/ t

= 10-5/10

=5/10

= 0.5m/s2

6 0

3 years ago

A race car drives one lap around a race track that is 500 meters in length.

pychu [463]

Explanation:

Check out the picture I drew for a minute before reading this...

B. Distance [the red line] is a scalar quantity reflecting how far an object has traveled. Displacement [the green line] is a vector quantity reflecting how far an object has moved from a point. The key difference is that distance can be any sort of path while displacement is always a vector (or a straight line) between a starting point and a finishing point. Sometimes distance and displacement are equal to one another. Sometimes you have a distance traveled, but zero displacement overall; which is what's going on in your question.

A. The distance that the racecar traveled is indeed 500m. But at the end of the lap, it is right back where it started. So overall, it has been displaced 0m.

3 0

3 years ago

You have a pendulum clock made from a uniform rod of mass M and length L pivoting around one end of the rod. Its frequency is 1

drek231 [11]

The new oscillation frequency of the pendulum clock is 1.14 rad/s.

The given parameters;

Mass of the pendulum, = M
Length of the pendulum, = L
Initial angular speed, $\omega _i$ = 1 rad/s

The moment of inertia of the rod about the end is given as;

$I_i = \frac{1}{3} ML^2$

The moment of inertia of the rod between the middle and the end is calculated as;

$I_f = \int\limits^L_{L/2} {r^2\frac{M}{L} } \, dr = \frac{M}{3L} [r^3]^L_{L/2} = \frac{M}{3L} [L^3 - \frac{L^3}{8} ] = \frac{M}{3L} [\frac{7L^3}{8} ]= \frac{7ML^2}{24}$

Apply the principle of conservation of angular momentum as shown below;

$I _i \omega _i = I _f \omega _f\\\\\frac{ML^2}{3} (1 \ rad/s)= \frac{7ML^2}{24} \times \omega _f\\\\\frac{24 \times ML^2}{3 \times 7 ML^2} (1 \ rad/s)= \omega _f\\\\1.14 \ rad/s = \omega _f$

Thus, the new oscillation frequency of the pendulum clock is 1.14 rad/s.

Learn more about moment of inertia of uniform rod here: brainly.com/question/15648129

3 0

2 years ago