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Llana [10]
3 years ago
10

The specific heat of granite is 800 J/(kg·°C). How much heat does it take to raise 1 kg of granite 4°C?

Physics
1 answer:
Triss [41]3 years ago
7 0
Q=mcΔt
Q= 1kg * 800J/kg°C*4°C
Q= 3200J
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On my science test, there is a bonus question that I want to get right. Why would it be a bad idea to skydive on the moon? Hint:
eduard

Answer:

Because there is no air resistance

Explanation:

When an object falls on Earth, there are essentially two forces acting on it:

- The force of gravity, downward, equal to the weight of the object:

W=mg

where m is the mass and g the acceleration due to gravity

- The air resistance, F, which acts upward, and whose magnitude is generally proportional to v, the speed of the object

When the object starts its fall, its initial speed is zero: v = 0, so the air resistance is also zero: F=0, and the object accelerates downward due to gravity.

However, as it accelerates downward, its speed increases, and so does the air resistance F. However, F is upward, opposite to the direction of motion, therefore it reduces the net acceleration of the object; at a certain point, the magnitude of the air resistance will become equal to the weight, so that

mg = F

and at that point, the net acceleration of the object will become zero: this means that the object will continue its fall at a constant velocity, called terminal velocity.

On the Moon instead, there is no air resistance: this means that for an object falling down, the speed keeps increasing due to the effect of gravity, and it will never reach a terminal value: therefore, the final velocity at the impact will be much higher than on the Earth, if we assume the two objects have been dropped from a very high altitude from the surface.

7 0
3 years ago
A quarterback is set up to throw the football to a receiver who is running with a constant velocity v⃗ rv→rv_r_vec directly away
Artist 52 [7]

Answer:

a) V_o,y = 0.5*g*t_c

b) V_o,x = D/t_c - v_r

c) V_o = sqrt ( (D/t_c - v_r)^2 + (0.5*g*t_c)^2)

d)  Q = arctan ( g*t_c^2 / 2*(D - v_r*t_c) )

Explanation:

Given:

- The velocity of quarterback before the throw = v_r

- The initial distance of receiver = r

- The final distance of receiver = D

- The time taken to catch the throw = t_c

- x(0) = y(0) = 0

Find:

a) Find V_o,y, the vertical component of the velocity of the ball when the quarterback releases it.  Express V_o,y in terms of t_c and g.

b) Find V_o,x, the initial horizontal component of velocity of the ball.   Express your answer for V_o,x in terms of D, t_c, and v_r.

c) Find the speed V_o with which the quarterback must throw the ball.  

   Answer in terms of D, t_c, v_r, and g.

d) Assuming that the quarterback throws the ball with speed V_o, find the angle Q above the horizontal at which he should throw it.

Solution:

- The vertical component of velocity V_o,y can be calculated using second kinematics equation of motion:

                               y = y(0) + V_o,y*t_c - 0.5*g*t_c^2

                              0 = 0 + V_o,y*t_c - 0.5*g*t_c^2

                               V_o,y = 0.5*g*t_c

- The horizontal component of velocity V_o,x witch which velocity is thrown can be calculated using second kinematics equation of motion:

- We know that V_i, x = V_o,x + v_r. Hence,

                               x = x(0) + V_i,x*t_c

                               D = 0 + V_i,x*t_c

                               V_o,x + v_r = D/t_c

                                V_o,x = D/t_c - v_r

- The speed with which the ball was thrown can be evaluated by finding the resultant of V_o,x and V_o,y components of velocity as follows:

                           V_o = sqrt ( V_o,x^2 + V_o,y^2)

                          V_o = sqrt ( (D/t_c - v_r)^2 + (0.5*g*t_c)^2)

       

- The angle with which it should be thrown can be evaluated by trigonometric relation:

                            tan(Q) = ( V_o,y / V_o,x )

                            tan(Q) = ( (0.5*g*t_c)/ (D/t_c - v_r) )

                                   Q = arctan ( g*t_c^2 / 2*(D - v_r*t_c) )

                           

                               

6 0
3 years ago
What gases can CFC and HCFC refrigerants decompose into at high temperatures
nekit [7.7K]

Answer:

Hydrochloric and Hydrofluoric Acids.

3 0
2 years ago
What is an unbalanced force
Butoxors [25]
There's no such thing as "an unbalanced force".

If all of the forces acting on an object all add up to zero, then we say that
<span>the group </span>of forces is balanced.  When that happens, the group of forces
has the same effect on the object as if there were no forces on it at all. 

An example: 
Two people with exactly equal strength are having a tug-of-war.  They pull
with equal force in opposite directions.  Each person is sweating and straining,
grunting and groaning, and exerting tremendous force.  But their forces add up
to zero, and the rope goes nowhere.  The <u>group</u> of forces on the rope is balanced.

On the other hand, if one of the offensive linemen is pulling on one end of
the rope, and one of the cheerleaders is pulling on the other end, then their
forces don't add up to zero, because even though they're opposite, they're
not equal.  The <u>group</u> of forces is <u>unbalanced</u>, and the rope moves.

A group of forces is either balanced or unbalanced.  A single force isn't.
7 0
3 years ago
Water is considered a polar molecule because _____.
Mashutka [201]
<span>The question is asking for the reason why water is considered a polar molecule. The correct option is A. Water molecule is a polar molecule because it has one side which is positively charged and the other side is negatively charged. Each molecule of water has two hydrogen atom and one oxygen atom. The slight negative charge near the oxygen atom attract nearby hydrogen atoms from water and the slightly positive charge of the hydrogen atom attract oxygen atom. The attraction set up by these charges are responsible for the polarity of water molecules. </span>
4 0
3 years ago
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