All categories

4 years ago

A cat with mass 4.0 kg jumps down to the floor from a bookcase 2.0 m high. What is the cat's change in

Physics

2 answers:

lara31 [8.8K]4 years ago

8 0

Answer:

-78 J

Explanation:

I just answered that question.

eduard4 years ago

3 0

Answer: -7J

Explanation:

To determine the change in potential energy, use the equation ΔPE=mgΔh

PE=mg

, where m is the mass, g= 9.8 m/s2

g= 9.8 m/s

, and Δh is the change in height. Hence, we have that the potential energy PE=4.0 kg×9.8 m/s2×-2.0 m=-78 J

PE=4.0 kg

9.8 m/s

-2.0 m=-78 J

. To verify that this is correct, note that since the cat changes the potential energy to kinetic energy by jumping, the potential energy decreases. Hence, the potential energy should be negative.

You might be interested in

A device for measuring atmospheric pressure is a _____. thermometer manometer barometer seismometer

vodomira [7]

A device used to measure atmospheric pressure is a barometer.

8 0

4 years ago

Read 2 more answers

Question 9

egoroff_w [7]

Answer:

Explanation:

F=ma

given solution

v=12m/s a=v/t

s=6 sec =12m/s÷6sec

=2m/s^2 then we get acceleration now we will find the mass. first derive the the formula of mass by crisis cross then you will get this formula which is m=F/a

=36÷2

= 18

6 0

3 years ago

You’ve made the finals of the science Olympics. As one of your tasks you’re given 1.0 g of copper and asked to make a cylindrica

Pani-rosa [81]

Answer:

Length = 2.92 m

Diameter = 0.11 mm

Explanation:

We have $m = dl D \ \ \& \ \ \ R = \frac{\rho l}{A}$ , where:

$l$ is the length

$m = 1.0 g = 1 \times 10^{-3} \ kg\\R = 1.3 \ \Omega\\\rho = 1.7 \times 10^{-8} \Omega m\\d = 8.96 \ g/cm^3 = 8960 kg/m^3$

We divide the first equation by the second equation to get:

$\frac{m}{R} = \frac{d A^2}{\rho}$

$A^2 = \frac{m \rho}{dR} \\\\A^2 = \frac { 1 \times 10^{-3} \times 1.7 \times 10^{-8}}{8960 \times 1.3}\\\\A^2 = 1.5 \times 10^{-15}\\\\ A= 3.8 \times 10^{-8} \ m^2$