How much work is done in holding a 10 newton sack of potatoes while waiting in line at the grocery store for 3 minutes.

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

NISA [10]

The work done by the shopping basket is 147 J.

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

6 0

2 years ago

guapka [62]

Although X-rays have many benefits, they can be dangerous to human health because they are short wavelength, high frequency, with photons sufficiently energetic to damage DNA molecules.

4 0

3 years ago

Read 2 more answers

Light rays in a material with index of refrection 1.35 1.35 can undergo total internal reflection when they strike the interface

nekit [7.7K]

Answer:

1.30

Explanation:

To calculate the critical angle we have ti use the formula:

$sin\theta_c=\frac{n_2}{n_1}$

where theta_c is the critical angle, n1 is the index of refraction of the material where the light is totally reflected, and n2 is the refractive index of the other material.

By taking n_2 and replacing we obtain:

$n_2=n_1sin\theta_c=(1.35)sin75.1\°=1.30$

hope this helps!!

6 0

3 years ago

if a Firebird travels at a velocity of 0 to 60 mph in four seconds traveling east what was the acceleration of the Firebird

Tresset [83]

Answer:

6.7 m/s^2

Explanation:

The formula of acceleration is:

$\displaystyle{\vec{a} = \dfrac{\Delta \vec{v}}{\Delta t} = \dfrac{v_2 - v_1}{t_2-t_1}}$

where $\displaystyle{\vec{a}}$ is acceleration, $\displaystyle{\vec{v}}$ is velocity and $\displaystyle{t}$ is time. $\displaystyle{v_2}$ means final velocity. $\displaystyle{v_1}$ means initial velocity, $\displaystyle{t_2}$ means final time and $\displaystyle{t_1}$ means initial time.

We are given that the Firebird travels at velocity of 0 to 60 mph in four seconds. Therefore:

Our initial velocity starts at 0 mph.
Our final velocity is at 60 mph.
Our initial time is 0 second.
Our final time is 4 seconds.

Since it travels to the east then our vector will be positive. However, acceleration has to be in m/s^2 unit (Sl unit) so we'll have to convert from mph (miles per hours) to m/s (meters per second) first.

We know that:

A mile equals to 1609.344 meters.
An hour equals to 60 minutes which a minute equals to 60 seconds. So 60 minutes will equal to 3600 seconds.

Now we divide 1609.344 by 3600 to find a unit rate of m/s:

$\displaystyle{\dfrac{1609.344}{3600} \ \, \sf{m/s}}\\\\\displaystyle{= 0.44704 \ \, \sf{m/s}}$

Now multiply 0.44704 m/s by 0 and 60 to get velocity in m/s unit:

Initial velocity = 0 m/s
Final velocity = 60 * 0.44704 = 26.82 m/s

Time is already in second so no need for conversion. Substitute known information in the formula:

$\displaystyle{\vec{a} = \dfrac{26.82-0}{4-0}}\\\\\displaystyle{\vec{a} = \dfrac{26.82}{4}}\\\\\displaystyle{\vec{a} = 6.7 \ \, \sf{m/s^2}}$

Therefore, the Firebird will accelerate at the rate of 6.7 m/s^2.

3 0

2 years ago

A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.

kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

$W=Fd cos \theta$

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

$\theta=30^{\circ}$ is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

$W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J$

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

$\Delta K = W$

where

$\Delta K$ is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

$\Delta K=69.3 J$

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

$\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$ (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, $\Delta K$ , after it has travelled for d=3 m. Using the work-energy theorem again, we find

$\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J$

And substituting into (1) and re-arrangin the equation, we find

$v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s$

6 0

4 years ago

How much work is done in holding a 10 newton sack of potatoes while waiting in line at the grocery store for 3 minutes.​

How much work is done in holding a 10 newton sack of potatoes while waiting in line at the grocery store for 3 minutes.