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Tasya [4]
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)
Physics
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

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Nimfa-mama [501]

Answer:

The answer is Dependent Variable

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2 years ago
When a container decreases in size, what will happen to an amount of gas in the container??
alexgriva [62]
The gas would also decrease in size since the container lost gas to decrease the size of the container.
7 0
3 years ago
a bal is launched upward with a velocity of v0 from the edge of a cliff of height D. it reaches a maximum height of H above its
lilavasa [31]

Answer:

D/H =15

Explanation:

  • We can find first the peak height H, taking into consideration, that at the maximum height, the ball will reach momentarily to a stop.
  • At this point, we can find the value of H, applying the following kinematic equation:

       v_{f} ^{2} -v_{0} ^{2} = 2* g* H (1)

  • If vf=0, if we assume that the positive direction is upwards, we can find the value of H as follows:

       H = \frac{v_{0} ^{2} }{2*g} (2)

  • We can use the same equation, to find the value of D, as follows:

        v_{f} ^{2} -v_{1} ^{2} = 2* g* D (3)

  • In order to find v₁, we can use the same kinematic equation that we used to get H, but now, we know that v₀ = 0.
  • When we replace these values in (1), we find that  v₁ = -v₀.
  • Replacing in (3), we have:

        (4*v_{0})^{2} - (-v_{0}) ^{2}  = 2* g* D\\ \\ 15*v_{0}^{2}  = 2*g*D

  • Solving for  D:

       D = \frac{15*v_{0} ^{2} }{2*g}

  • From (2) we know that H can be expressed as follows:

       H = \frac{v_{0} ^{2} }{2*g}

  • ⇒ D = 15 * H

        \frac{D}{H} = 15

3 0
2 years ago
The potential difference between the surface of a 3.0-cm-diameter power line and a point 1.0 m distant is 3.9 kV. Find the line
Korolek [52]

Answer:

The linear charge density is 5.19 X 10⁻⁶ C/m

Explanation:

The potential difference between two cylinders, is given as

V = (λ/2πε)ln(b/a)

where;

λ is the line charge density on the power line.

b is the distance between the power line = 1 m

a is the radius of the wire = 1.5 cm = 0.015 m

ε is the permittivity of free space = 8.9 X 10⁻¹² C

V*2πε = λ* ln(b/a)

3900 *(2π*8.9 x10⁻¹²)= λ *ln(1/0.015)

2.1812 X 10⁻⁷ = 4.1997* λ

λ = 5.19 X 10⁻⁶ C/m

Therefore, the linear charge density is 5.19 X 10⁻⁶ C/m

6 0
2 years ago
a bullet is fired horizontally with a speed of 524 m/s from a height of 22m above the ground. calculate where it will hit the gr
lara31 [8.8K]

Given data:

* The height of the bullet is 22 m.

* The speed of bullet in the horizontal direction is 524 m/s.

Solution:

By the kinematics equation, the time taken by the bullet to reach the ground is,

h=u_yt+\frac{1}{2}gt^2

where u_y is the vertical velocity component, t is the time taken to reach the ground, g is the acceleration due to gravity, and h is the height of the bullet,

Substituting the known values,

\begin{gathered} 22=0+\frac{1}{2}\times9.8\times t^2 \\ t^2=\frac{22\times2}{9.8} \\ t^2=4.49 \\ t=2.12\text{ s} \end{gathered}

Thus, the time taken by the bullet to reach the ground is 2.12 seconds.

By the kinematics equation for the horizontal motion, the horizontal range of the bullet is,

R=u_xt+\frac{1}{2}at^2

where u_x is the horizontal component of the velocity, a is the acceleration along the horizontal direction, t is the time taken to reach the ground and R is the horizontal range,

Substituting the known values,

\begin{gathered} R=524\times2.12+0 \\ R=1110.9\text{ m} \end{gathered}

Thus, the horizontal range of the bullet is 1110.9 meters.

Hence, the bullet hit the ground at 1110.9 meters.

8 0
11 months ago
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