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Mekhanik [1.2K]
2 years ago
12

Please I need help! This is the last question I need for this assignment!

Physics
1 answer:
Ne4ueva [31]2 years ago
6 0

Answer:

When the same amount of heat is added to cold sand and cold water, the temperature change of sand will be higher because of its lower specific heat capacity.

What is specific heat capacity?

Specific heat capacity is the quantity

of heat required to raise a unit mass of

a substance by 1 kelvin.

Specific heat capacity of water and sand

{<em>refer to the above attachment}</em>

Δθ = Q/mc

Thus, for an equal mass of water and sand, when the same amount of heat is added to cold sand and cold water, the temperature change of sand will be higher because of its lower specific heat capacity.

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An object traveling around another object in space is in
drek231 [11]
It is orbiting the object.
8 0
4 years ago
A student stands on the edge of a cliff that is 300 m high and kicks a rock horizontally. 7.8 seconds later, the rock hits the g
mart [117]

Answer:

6.65m/s

Explanation:

Using the equation of motion

S = ut + 1/2gt²

S is the height of fall

t is the time

u is the horizontal velocity

g is the acceleration due to gravity

Given

S = 300 + 50

S = 350m

t = 7.8seconds

g = 9.8m/s^2

Get S

S = 7.8u + 1/2(9.8)(7.8)²

S = 7.8u + 298.116

350 = 7.8u + 298.116

7.8u = 350 - 298.116

7.8u = 51.884

u = 51.884/7.8

u = 6.65m/s

Hence the rock's horizontal velocity was 6.65m/s

6 0
3 years ago
Cannonball a is launched across level ground with speed v₀ at an angle of 45. Cannonball b is launched from the same location wi
postnew [5]

Answer:

Cannonball b spends more time in the air than cannonball a.

Explanation:

Starting with the definition of acceleration, we have that:

a=\frac{\Delta v}{\Delta t}\\\\\Delta t= \frac{\Delta v}{a}

Since both cannonballs will stop in their maximum height, their final velocity is zero. And since the acceleration in the y-axis is g, we have:

\Delta t= -\frac{v_{oy}}{g}

Now, this time interval is from the moment the cannonballs are launched to the moment of their maximum height, exactly the half of their time in the air. So their flying time t_f is (the minus sign is ignored since we are interested in the magnitudes only):

t_f=2\frac{v_{oy}}{g}

Then, we can see that the time the cannonballs spend in the air is proportional to the vertical component of the initial velocity. And we know that:

v_{oy}=v_o\sin\theta\\\\\implies t_f=2\frac{v_o\sin\theta}{g}

Finally, since \sin60\°=\frac{\sqrt{3} }{2} and \sin45\°=\frac{\sqrt{2} }{2}, we can conclude that:

t_{fa}=\sqrt{2}\frac{v_o }{g} \\\\t_{fb}=\sqrt{3}\frac{v_o }{g}\\\\\implies t_{fb}>t_{fa}

In words, the cannonball b spends more time in the air than cannonball a.

5 0
3 years ago
A 10,000kg space ship is orbiting the moon with a radius of 25km from the ship to the center of the moon. It's tangential speed
galina1969 [7]

Answer:

False

Explanation:

ac = v^2/r

acceleration is not dependent on the mass of the orbiting object.

3 0
3 years ago
Problem: At the local swimming hole, a favorite trick is torun
denis23 [38]

Answer:

2 revolutions

Explanation:

Assume that when she runs off the edge of the 8.3m high cliff, her vertical speed is 0. So gravitational acceleration g = 9.8m/s2 is the only thing that makes her fall down. So we can use the following equation of motion to calculate the time it takes for her to fall down:

s = gt^2/2

where s = 8.3 m is the distance that she falls, t is the time it takes to fall, which is what we are looking for

t^2 = \frac{2s}{g} = \frac{2*8.3}{9.8} = 1.694

t = \sqrt{1.694} = 1.3 s

Since she rotates with an average angular speed of 1.6rev/s. The number of revolutions she would make within 1.3s is

rev = 1.3 * 1.6 = 2 revolution

8 0
4 years ago
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