Movement of a stationary object (2words)

A 700 g can of beans is dropped from a shelf that is 1.5 m high. What is the gravitational potential energy of this can? Round y

OLga [1]

<span>To find the gravitational potential energy of an object, we can use this equation: GPE = mgh m is the mass of the object in kg g = 9.80 m/s^2 h is the height of the object in meters GPE = mgh GPE = (0.700 kg) (9.80 m/s^2) (1.5 m) GPE = 10.3 J The gravitational potential energy of this can is 10.3 J</span>

4 0

3 years ago

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Arrange the stars based on their temperature. Begin with the coolest star, and end with the hottest star.

Slav-nsk [51]

Answer:

Uranus, Pluto, Neptune, Saturn , Jupiter, mars, Venus ,mercury and sun

4 0

2 years ago

A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters

Artist 52 [7]

Answer:

Part a)

$F = 7.76 N$

Part b)

$\theta = -137.7 degree$

Part c)

$\theta = -127.7 degree$

Explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

$x = -19 + t - 3t^3$

$y = 20 + 7t - 9t^2$

Part a)

differentiate x and y two times with respect to time to find the acceleration

$a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)$

$a_x = \frac{d}{dt}(0 +1 - 9t^2)$

$a_x = -18t$

$a_y = \frac{d^2}{dt^2}(20 + 7t - 9t^2)$

$a_y = \frac{d}{dt}(0 +7 - 18t)$

$a_y = -18$

Now the acceleration of the object is given as

$\vec a = (-18t)\hat i + (-18)\hat j$

at t= 1.1 s we have

$\vec a = -19.8 \hat i - 18 \hat j$

now the net force of the object is given as

$\vec F = m\vec a$

$\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)$

$\vec F = -5.74 \hat i - 5.22 \hat j$

now magnitude of the force will be

$F = \sqrt{5.74^2 + 5.22^2} = 7.76 N$

Part b)

Direction of the force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{-5.22}{-5.74}$

$\theta = -137.7 degree$

Part c)

For velocity of the particle we have

$v_x = \frac{dx}[dt}$

$v_x = (0 +1 - 9t^2)$

$v_y = \frac{dy}{dt}$

$v_y = (0 +7 - 18t)$

now at t = 1.1 s

$\vec v = -9.89\hat i - 12.8 \hat j$

now the direction of the velocity is given as

$\theta = tan^{-1}(\frac{v_y}{v_x})$

$\theta = tan^{-1}(\frac{-12.8}{-9.89})$

$\theta = -127.7 degree$

7 0

3 years ago

Please HELP ME If the motion of an object changes, what must be true about the forces acting on that object?

atroni [7]

Answer:

If there is a net force acting on an object, the object will have an acceleration and the object's velocity will change. ... Newton's second law states that for a particular force, the acceleration of an object is proportional to the net force and inversely proportional to the mass of the object.

Explanation:

4 0

3 years ago

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Convert the following to relative uncertainties <br>a) 2.70 ± 0.05cm<br>b) 12.02 ± 0.08cm

DENIUS [597]

data which is expressed in form of following way

$a = a_o + \Delta a$