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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.

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This subatomic particle adds the least amount of mass to an atom and is called the

sergey [27]

A nuetron is the lightest subatomic particle

8 0

3 years ago

You blow across the open mouth of an empty test tube and producethe fundamental standing wave of the air column inside the testt

sertanlavr [38]

Answer:

(a). The frequency of this standing wave is 0.782 kHz.

(b). The frequency of the fundamental standing wave in the air is 1.563 kHz.

Explanation:

Given that,

Length of tube = 11.0 cm

(a). We need to calculate the frequency of this standing wave

Using formula of fundamental frequency

$f_{1}=\dfrac{v}{4l}$

Put the value into the formula

$f_{1}=\dfrac{344}{4\times0.11}$

$f_{1}=781.81\ Hz$

$f_{1}=0.782\ kHz$

(b). If the test tube is half filled with water

When the tube is half filled the effective length of the tube is halved

We need to calculate the frequency

Using formula of fundamental frequency of the fundamental standing wave in the air

$f_{1}=\dfrac{v}{4(\dfrac{L}{2})}$

Put the value into the formula

$f_{1}=\dfrac{344}{4\times\dfrac{0.11}{2}}$

$f_{1}=1563.63\ Hz$

$f_{1}=1.563\ kHz$

Hence, (a). The frequency of this standing wave is 0.782 kHz.

(b). The frequency of the fundamental standing wave in the air is 1.563 kHz.

6 0

4 years ago

A particle undergoes two displacements. The first has a magnitude of 11 m and makes an angle of 82 ◦ with the positive x axis. T

Luden [163]

Answer:

$\theta = 211.7 degree$

Explanation:

First displacement of the particle is given as

$r_1$ = 11 m at 82 degree with positive X axis

so we can say

$\vec r_1 = 11 cos82 \hat i + 11 sin82 \hat j$

$\vec r_1 = 1.53\hat i + 10.9 \hat j$

resultant displacement of the particle after second displacement is given as

r = 8.7 m at 135 degree with positive X axis

so we can say

$r = 8.7 cos135\hat i + 8.7 sin135\hat j$

$r = -6.15 \hat i + 6.15 \hat j$

now we know that

$r = r_1 + r_2$

now we have

$r_2 = r - r_1$

so we will have

$r_2 = (-6.15 \hat i + 6.15 \hat j) - (1.53\hat i + 10.9 \hat j)$

$r_2 = -7.68 \hat i - 4.75 \hat j$

so angle of the second displacement is given as

$tan\theta = \frac{r_y}{r_x}$

$tan\theta = \frac{-4.75}{-7.68}$

$\theta = 211.7 degree$

8 0

3 years ago

On a winter day with a temperature of -10°C, 500g of snow (water ice) is brought inside where the temperature is 18 °C. The snow

Serjik [45]

Answer:

Explanation:

Mass of ice m = 500g = .5 kg

Heat required to raise the temperature of ice by 10 degree

= mass of ice x specific heat of ice x change in temperature

= .5 x 2093 x 10 J

10465 J

Heat required to melt the ice

= mass of ice x latent heat

0.5 x 334 x 10³ J

167000 J

Heat required to raise its temperature to 18 degree

= mass x specific heat of water x rise in temperature

= .5 x 4182 x 18

=37638 J

Total heat

=10465 +167000+ 37638

=215103 J

7 0

3 years ago

You and your friend Peter are putting new shingles on a roof pitched at 21∘. You're sitting on the very top of the roof when Pet

Marat540 [252]

Answer:

3.69 m/s

Explanation:

Forces :

mgsin Θ - mumgcosΘ = ma

g x sinΘ - mu x g x cosΘ = a

9.8 x sin 21 - 0.53 x 9.8 x cos 21 = a

a = -1.337 m/s²

so you have final velocity = 0 m/s

initial velocity = ? m/s

Given d = 5.1 m

By kinematics

vf² = vo² + 2ad

0 = vo² + 2 x -1.337*5.1

vo = 3.69 m/s

8 0

4 years ago