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2 years ago

A ball of 0.5kg slows down from 5m/s to 3m/s. Calculate the work done inthe process.

Physics

1 answer:

dlinn [17]2 years ago

5 0

Answer:

9.8 Joules (rounded to 2 significant figures)

Explanation:

Work done (J)= Force(N) x distance changed (m)

Force= 9.80665 x 0.5kg
Force= 4.90332 Newtons

Distance changed= 5-3
distance changed= 2m/s

--> work done= 4.90332 x 2

work done= 9.8 Joules

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A 1kg mass is thrown to a height of 2cm. what is the potential energy

prohojiy [21]

Mass=m=1kg
Height=h=2cm=0.02m
Acceleration due to gravity=g=10m/s^2

$\\ \tt\hookrightarrow P.E=mgh$

$\\ \tt\hookrightarrow PE=1(10)(0.02)$

$\\ \tt\hookrightarrow PE=0.2J$

4 0

1 year ago

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A volumen constante un gas ejerce una presión de 880 mmHg a 20o Celsius dentro de una olla a presión ¿Qué temperatura habrá si e

jasenka [17]

Answer: Hence, the final temperature is 350 K

Explanation :

To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=880mmHg\\T_1=20^0C=(20+273)K=293K\\P_2=1050mmHg\\T_2=?$

Putting values in above equation, we get:

$\frac{880mmHg}{293K}=\frac{1050mmHg}{T_2}\\\\T_2=350K$

Hence, the final temperature is 350 K

8 0

2 years ago

A long, East-West-oriented power cable carrying an

Alla [95]

Answer:

200A

Explanation:

Given that

the distance between earth surface and power cable d = 8m

when the current is flowing through cable , the magnitude flux density at the surface is 15μT

when the current flow throught is zero the magnitude flux density at the surface is 20μT

The change in flux density due to the current flowing in the power cable is

B = 20μT - 15μT

B =5μT -----(1)

The expression of magnitude flux density produced by the current carrying cable is

$B=\frac{\mu_0I}{2\pi d}$ -----(2)

Substitute the value of flux density

B from eqn 1 and eqn 2

$\frac{\mu_0I}{2\pi d}=5\times 10^-^6\\\\\frac{(4\pi \times 10^-^7)I}{2 \pi (8)} =5\times 10^-^6\\\\I=200A$

Therefore, the magnitude of current I is 200A

8 0

2 years ago

a bus starting from rest moves with a uniform acceleration of 0.1 metre per second square for 2 minutes find the speed acquired

blondinia [14]

S ?
U 0m/s
V ?
A 0.1m/s^2
T 2min (120 sec)

S=ut+0.5at^2
S=0(120 sec)+0.5(0.1m/s^2)(120 sec)^2
S=720m

Distance double 720m*2=1440m

V^2=u^2+2as
V^2=(0)^2+2(0.1 m/s^2)(1440m)
V^2=288
V= square root of 288=12 root 2=16.97 to 2 decimal places

6 0

2 years ago

A mass of 30.0 grams hangs at rest from the lower end of a long vertical spring. You add different amounts of additional mass ΔM

ra1l [238]

Answer:

K = 373.13 N/m

Explanation:

The force of the spring is equals to:

Fe - m*g = 0 => Fe = m*g

Using Hook's law:

K*X = m*g Solving for K:

K = m/X * g

In this equation, m/X is the inverse of the given slope. So, using this value we can calculate the spring's constant:

K = 10 / 0.0268 = 373.13N/m

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3 years ago

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