Ask question

Ask question

All categories

2 years ago

15

Ahmed fills a basket with shopping weighing 10 kg. How much work is being done on the shopping basket when he lifts it verticall

y by 1.5 m? (g= 9.8 m/s²)

1 answer:

NISA [10]2 years ago

6 0

The work done by the shopping basket is 147 J.

<h3>When is work said to be done?</h3>

Work is said to be done whenever a force moves an object through a certain distance.

The amount of work done on the shopping basket can be calculated using the formula below.

Formula:

W = mgh

Where:

W = Amount of work done by the basket
m = mass of the shopping basket
h = height of the shopping basket
g = acceleration due to gravity.

Form the question,

Given:

m = 10 kg
h = 1.5 m
g = 9.8 m/s²

Substitute these values into equation 2

W = 10(1.5)(9.8)
W = 147 J.

Hence, The work done by the shopping basket is 147 J.

Learn more about work done here: brainly.com/question/18762601

You might be interested in

Acids turned _______bases turned _________and neutral stayed______

garik1379 [7]

I'm not sure tbh, but if it helps, heres a game that helped me understand bases acids and neutrals for chem in 8th grade! http://static.lawrencehallofscience.org/scienceview/scienceview.berkeley.edu/html/showcase/flash/jui... I know it looks cheesy, but it helps! Chem is something youll have until college, and it helps to learn about it. You can make drinks using acids, bases, and neutrals!

8 0

3 years ago

The current that charges a capacitor transfers energy that is stored in the capacitor’s electric field. Consider a 2.0 μF capaci

lapo4ka [179]

Answer:

the capacitor voltage is V = 20 V

Explanation:

Given,

Capacitance of the capacitor = 2.0 μF

energy stored = 200 W

time (t) =2.0 μs

Capacitor voltage = ?

$\dfrac{dE}{dt}=200 W$

$E =200\times 2 \times 10^{-6}$

$E =400 \times 10^{-6}\ J$

we know,

$E = \dfrac{1}{2}CV^2$

$V =\sqrt{\dfrac{2E}{C}}$

$V =\sqrt{\dfrac{2\times 400 \times 10^{-6}}{2\times 10^{-6}}}$

$V =\sqrt{400}$

V = 20 V

so , the capacitor voltage is V = 20 V

7 0

3 years ago

Ilia_Sergeevich [38]

Answer:

The value is $C = 30729\ c$

Explanation:

From the question we are told that

The power rating of the stove is $P = 4.4 KW = 4.4 *10^{3} \ W$

The duration of its use everyday is $t_1 = 70 \ minutes = 1.167 \ hours$

The rating of the light bulbs is $P_2' = 150 W$

The number is $n = 7$

The power rating of the total bulb is $P_2 = 7 * 150 = 1050 \ W$

The duration of its use everyday is $t_2 = 7 hours$

The power rating of miscellaneous appliance $P_3 = 1.8 \ KW = 1.8 *10^{3} \ W$

The duration of its use everyday is $t_3 = 1 hour$

The power rating of hot water $P = 4 KW = 4 *10^{3} \ W$

The duration of its use everyday is $t_4 = 120 \ minutes = 2 hours$

Generally the total electrical energy used in 1 month is mathematically represented as

$E = P_1 * t_1 * 30 + P_2 * t_2 * 30 + P_3 * t_3 * 30+ P_4 * t_4 * 30$

=> $E = 4.4*10^{3} * 1.167 * 30 + 1050 * 7 * 30 + 1.8 *1 * 30+ 4*10^{3} * 2 * 30$

=> $E = 614598 \ W \cdot h$

=> $E = 614.6 \ K W \cdot h$

Generally the monthly electricity bill is mathematically represented as

$C = 614.6 * 50$

=> $C = 30729\ c$

8 0

2 years ago

A ball falls from the top of a building. As it falls, its speed increases. Which type of energy is the ball gaining as it falls?

Elina [12.6K]

<h2>Hello!</h2>

The answer is: B. Kinetic energy

<h2>Why?</h2>

Since the ball is falling, speed increases because the gravity acceleration is acting. When speed increases, the kinetic energy increases too, so the ball is gaining kinetic energy.

The gravity acceleration is equal to $9.81\frac{m}{s^{2}}$ , it means that when falling, the ball will increase it's speed 9.81m every second.

We can calculate the kinetic energy by using the following formula:

$KE=\frac{1}{2}*m*v^{2}$

Where:

$m=mass\\v=velocity$

Have a nice day!

<h2 />

5 0

3 years ago

An object is launched with an initial velocity of 50.0 m/s at a launch angle of 36.9∘ above the horizontal. part a determine x-v

ankoles [38]

Given:
v = 50.0 m/s, the launch velocity
θ = 36.9°, the launch angle above the horizontal

Assume g = 9.8 m/s² and ignore air resistance.
The vertical component of the launch velocity is
Vy = (50 m/s)*sin(50°) = 30.02 m/s

The time, t, to reach maximum height is given by
(30.02 m/s) - (9.8 m/s²)*(t s) = 0
t = 3.0634 s
The time fo flight is 2*t = 6.1268 s

The horizontal velocity is
u = (50 m/s)cos(36.9°) = 39.9842 m/s
The horizontal distance traveled at time t is given in the table below.

Answer:

t, s x, m
------ --------
0 0
1 39.98
2 79.79
3 112.68
4 159.58
5 199.47
6 239.37

5 0

3 years ago

Read 2 more answers