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2 years ago

7

A spring with a spring constant of 2.5 N/m is stretched by 2

1 answer:

mr_godi [17]2 years ago

3 0

Answer:

Eelas = 5 [J]

Explanation:

Elastic energy is associated with the ability that has a spring or any other material with elastic properties.

We can calculate the elastic energy by means of the following equation:

$E_{elas}=\frac{1}{2} *k*x^{2}$

where:

k = constant spring = 2.5 [N/m]

x = distance stretched = 2 [m]

$E_{elas}=\frac{1}{2} *2.5*(2)^{2}\\E_{elas}=5[J]$

You might be interested in

A 1300 kg car traveling with a speed of 3.5 m/s executes a turn with a 8.5 m radius of curvature.

Y_Kistochka [10]

Answer:

1.4 m/s/s (2.s.f)

Explanation:

The formula for centripetal acceleration is:

$a=\frac{v^{2} }{r}$ , where v is velocity and r is the radius.

In the question we are given the information that the car has a mass of 1300kg, a velocity of 2.5m/s, and a turn radius of 8.5m which are all the values we need. Therefore we can simply substitute in the values to solve the question:

$a=\frac{3.5^{2} }{8.5} \\a=1.4$

Therefore the centripetal acceleration of the car is 1.4m/s/s. (2.s.f)

Hope this helped!

7 0

2 years ago

In boxing, the use of 16-ounce gloves rather than 12-ounce gloves reduces the chance of injury because the force is distributed

mr Goodwill [35]

Answer:

true

Explanation:

Here we have assumed that increasing the mass of a glove will increase the surface area.

Injury is caused by the application of pressure at a point on the body. The application of pressure takes place via the area of the gloves. Pressure is given by

$P=\dfrac{F}{A}$

where

F = Force

A = Area to which the force is applied

So, a bigger glove will increase the surface area and reduce the pressure resulting in a lower chance of injury.

Hence, the statement is true.

5 0

2 years ago

Citrus2011 [14]

Answer:

net force is positive downward..B

6 0

2 years ago

An incompressible fluid flows steadily through a pipe that has a change in diameter. The fluid speed at a location where the pip

OverLord2011 [107]

Answer:

The value is $v_2 = 5.53 \ m /s$

Explanation:

From the question we are told

The pipe diameter at location 1 is $d = 8.8 \ cm = \frac{8.8 }{10} = 0.88 \ m$

The velocity at location 1 is $v_1 = 2.4 \ m /s$

The diameter at location 2 is $d_2 = 5.80 \ cm = 0.58 \ m$

Generally the area at location 1 is

$A_1 = \pi * \frac{d^2}{ 2}$

=> $A_1 = \pi * \frac{0.88^2}{ 2}$

=> $A_1 = 3.142 * \frac{0.88^2}{ 2}$

=> $A_1 = 1.2166 \ m^2$

Generally the area at location 1 is

$A_2 = \pi * \frac{d_1^2}{ 2}$

=> $A_2 = \pi * \frac{0.58^2}{ 2}$

=> $A_2 = 0.528 \ m^2$

Generally from continuity equation we have that

$A_1 * v_1 = A_2 * v_2$

=> $1.2166 * 2.4 = 0.528 * v_2$

=> $1.2166 * 2.4 = 0.528 * v_2$

=> $v_2 = 5.53 \ m /s$

3 0

2 years ago

A rectangular barge, 5.2 m long and 2.4 m wide, floats in fresh water. Suppose that a 410-kg crate of auto parts is loaded onto

sesenic [268]

<h2>Answer:</h2><h2>The depth of barge float=3 cm</h2><h2>Explanation:</h2>

Length of rectangular barge=5.2 m

Width of rectangular barge=2.4m

Mass of crate=410 kg

Let h be the height of barge float

Volume of barge float= $l\times b\times h=5.2\times 2.4\times h=12.48h$

Density of water= $10^3kg/m^3$

Weight of water displaced by barge=Buoyant force=-Weight of horse

$Volume\;of\;water\times density\;of\;water\times g=410\times g$

$12.48h\times 1000=410$

$h=\frac{410}{12.48\times 1000}=0.03 m$

1 m=100 cm

$0.03 m=0.03\times 100=3$ cm

Hence, the depth of barge float=3 cm

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4 0

3 years ago

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