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2 years ago

8

Roughly to what height would a 5 kg stone need to be raised in order to have the same stored energy as the energy stored in the

defibrillator capacitor?

1 answer:

hoa [83]2 years ago

8 0

8.16m is the required height, a 5kg stone need to be raised.

One sort of potential energy is gravitational potential energy, which is equal to the product of the object's mass (m), the gravitational acceleration (g), and the object's height (h) as measured in relation to the ground's surface (the body).

We obtain the formula by considering the work done in raising a mass m through a height h.

Work in elevating mass m through height h is equal to force times distance.

The force must be greater than the mass m's weight, hence F = mg.

Work done = mgh = gravitational potential energy

Energy = Mass of the object × gravitational acceleration × height.

Mass of the stone = 5kg

Equating ;

∴ 400 J = 5 kg × 9.8 m/s² × height

Height = 8.16 m

Therefore, 8.16m is the required height.

Learn more about energy here:

brainly.com/question/1242059

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Please please help me please please help please please help me please

Kaylis [27]

Answer:

Electromagnetic waves do not require any medium to travel whereas mechanical waves must have a medium to propagate.

So, Basically, it is B I believe.

Hope It Helps!

6 0

2 years ago

An artillery shell is fired with an initial velocity of 300 m/s at 52.0° above the horizontal. To clear an avalanche, it explode

sasho [114]

The x- and y-coordinates are 9142.57 m and -304.425 m

<u>Explanation:</u>

As the motion of the shell is in a plane (two dimensional space) and the acceleration is that due to gravity which is vertically downward, we resolve initial velocity of the shell $v_{0}$ in horizontal and vertical directions. If the initial velocity of the shell is making angle with the horizontal, the horizontal component of initial velocity will be

$v_{x}=v_{0} \times \cos \theta$

As the acceleration of the shell is vertical having no horizontal component, the shell may be considered to move horizontally with constant velocity of $v_{x}$ and hence the horizontal distance covered (or the x coordinate of the shell with point of projection as origin) is given by

$v_{x}=v_{o} \times \cos \theta=300 \times \cos \left(52^{\circ}\right)=184.69 \mathrm{m} / \mathrm{s}$

$v_{y}=v_{o} \times \sin \theta==300 \times \sin \left(52^{\circ}\right)=236.4 \mathrm{m} / \mathrm{s}$

For motion with constant acceleration, we know

$s=s_{0}+v_{0} t+\left(\frac{(1)}{2}\right) a t^{2}$

Along the horizontal, x-axis, we might write this as

$x=x_{0}+v_{x 0} t+\left(\frac{1}{2}\right) a_{x} t^{2}$

Measuring distances relative to the firing point means

$x_{0}=0$

we know that,

$a_{x}=0$

or,

$v_{x}=v_{x 0}=\text { constant }$

By applying the values, we get,

$x=0+(184.69 \times 49.5)+\left(\left(\frac{1}{2}\right) \times 0 \times(49.5)^{2}\right)=9142.57 \mathrm{m}$

The acceleration of gravity is vertically downward and is $g=-9.8 \mathrm{m} / \mathrm{s}^{2}$ , hence the vertical distance covered (or y coordinate of the shell) is given by the second equation of motion

$y=y_{0}+v_{y 0} t+\left(\frac{1}{2}\right) a_{y} t^{2}$

we know, $y_{0}=0$ and $a_{y}=-9.8 \mathrm{m} / \mathrm{s}^{2}$ , so,

$y=0+(236.4 \times 49.5)+\left(\left(\frac{1}{2}\right) \times(-9.8) \times(49.5)^{2}\right)$

y = 11701.8 - 4.9(2450.25)= 11701.8 - 12006.225 = - 304.425 m

7 0

3 years ago

A 50kg car is moving at a velocity of 4 m/s. After a collision, the car is moving at a velocity of 5m/s.

earnstyle [38]

Answer:

Explanation:

The equation for momentum is p = mv; therefore,

p1 = 50(4) so

p1 = 200

p2 = 50(5) so

p2 = 250

The impulse, the change in the momentum, is 50, going in the same direction.

8 0

2 years ago

Which of the following statements best describes the function of the muscular system?

Korvikt [17]

Explanation:

A is excretion

B is digestion

so my guess is c

8 0

3 years ago

A can of beans that has mass M is launched by a spring-powered device from level ground. The can is launched at an angle of α0 a

scZoUnD [109]

Hi there!

A.

Since the can was launched from ground level, we know that its trajectory forms a symmetrical, parabolic shape. In other words, the time taken for the can to reach the top is the same as the time it takes to fall down.

Thus, the time to its highest point:
$T_h = \frac{T}{2}$

Now, we can determine the velocity at which the can was launched at using the following equation:
$v_f = v_i + at$

In this instance, we are going to look at the VERTICAL component of the velocity, since at the top of the trajectory, the vertical velocity = 0 m/s.

Therefore:
$0 = v_y + at\\\\0 = vsin\theta - g\frac{T}{2}$

***vsinθ is the vertical component of the velocity.

Solve for 'v':
$vsin(\alpha_0) = g\frac{T}{2}\\\\v = \frac{gT}{2sin(\alpha_0)}$

Now, recall that:
$W = \Delta KE = \frac{1}{2}m(\Delta v)^2$

Plug in the expression for velocity:
$W = \frac{1}{2}M (\frac{gT}{2sin(\alpha_0)})^2\\\\\boxed{W = \frac{Mg^2T^2}{8sin^2(\alpha _0)}}$

B.

We can use the same process as above, where T' = 2T and Th = T.

$v = \frac{gT}{sin(\alpha _0)} }\\\\W = \frac{1}{2}M(\frac{gT}{sin(\alpha _0)})^2\\\\\boxed{W = \frac{Mg^2T^2}{2sin^2(\alpha _0)}}$

C.

The work done in part B is 4 times greater than the work done in part A.

$\boxed{\frac{W_B}{W_A} = \frac{4}{1} = 4.0}$

4 0

2 years ago