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crimeas [40]
2 years ago
11

If you drop a 50 gram piece of metal that has a temperature of 110°Celsius into 1000 grams of water at 25°Celsius, what best des

cribes what would occur?
A.) The water will quickly reach the boiling point.
B.) The water’s temperature will stay the same, but the metal will cool down.
C.) The water’s temperature will increase, and the metal will stay constant.
D.)The water and the metal’s temperature will reach the same temperature.
Physics
2 answers:
zaharov [31]2 years ago
8 0

Answer:

D.)The water and the metal’s temperature will reach the same temperature.

Explanation:

As we know that heat is a dynamic energy which will flow from high temperature to low temperature.

So here we know that 50 g metal piece at 110 degree Celsius is dropped into the water at 25 degree Celsius. Since metal is at higher temperature so here the heat will flow from metal to the water and this will continue till the temperature of these two bodies will become equal.

So here the final temperature of metal and the water will be same at thermal equilibrium.

So here correct answer will be

D.)The water and the metal’s temperature will reach the same temperature.

LuckyWell [14K]2 years ago
7 0
If you drop a 50 gram piece of metal that has a temperature of 110°Celsius into 1000 grams of water at 25°Celsius, <span>D.)The water and the metal’s temperature will reach the same temperature. In any system undergoing heat transfer, the objects involved will eventually reach the same temperature, signifying thermal equilibrium.</span>
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A radar receiver indicates that a pulse return as an echo in 20 μs after it was sent. How far away is the reflecting object? (c
Assoli18 [71]

A radar receiver indicates that a pulse return as an echo in 20 μs after it was sent. The reflecting object would be 3000 m away .

Phenomenon of hearing back our own sound is called an echo. It is due to successive reflection of sound waves from the surfaces or obstacles of large size. To hear an echo, there must be a time gap of 0.1 second in original sound and the reflected sound.

Given

time =  20 μs = 20 * 10^{-6} s

let distance to the reflecting surface be = x

total distance travelled by pulse will be  = 2x

speed = 3.0 × 10^{8} m/s

distance = speed * time

2x = 3.0 × 10^{8} * 20 * 10^{-6}

   x = 3000 m

The reflecting object would be 3000 m away

To learn more about echo here

brainly.com/question/14861578?referrer=searchResults

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5 0
1 year ago
What is the potential energy for and object with mass of 15 Kg and 20 m above the ground?
vaieri [72.5K]

Answer:

3000J

Explanation:

The gravitational energy for any object is given by E=mgh

where m is mass, g is gravitational field strength anf h is height above ground

5 0
3 years ago
If a baseball's velocity is increased to four times its original velocity, by what factor does its kinetic energy increase
jeyben [28]

Answer:

16 times

Explanation:

Original

KE(i)  = 1/2mv²

After increasing

KE(f) = 1/2m(4v)²

KE(f) = 8mv²

KE(f)/KE(i) = 8 x 2 = 16

8 0
2 years ago
A student pulls a block over a rough surface with a constant force FP that is at an angle θ above the horizontal, as shown above
gizmo_the_mogwai [7]

Answer:

B.The force of friction between the block and surface will decrease.

Explanation:

The force of friction is given by

F_s = \mu N

where \mu is the coefficient of friction and N is the normal force.

When the student pulls on the block with force F_p at an angle \theta, the normal force on the block becomes

N  = Mg- F_psin(\theta)

and hence the frictional force becomes

F_s = \mu (Mg- F_psin(\theta)).

Now, as we increase \theta, sin(\theta) increases which as a result decreases the normal force Mg- F_psin(\theta), which also means the frictional force decreases; Hence choice B stands true.

<em>P.S: Choice D is tempting but incorrect since the weight </em>W=mg<em> is independent of the external forces on the block. </em>

6 0
3 years ago
Read 2 more answers
A kangaroo can jump over an object 2.46 m high. (a) Calculate its vertical speed (in m/s) when it leaves the ground.
Elenna [48]

Answer:

a) 6.95 m/s

b) 1.42 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 2.46-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 2.46}\\\Rightarrow u=6.95\ m/s

a) The vertical speed when it leaves the ground. is 6.95 m/s

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-6.95}{-9.81}\\\Rightarrow t=0.71\ s

Time taken to reach the maximum height is 0.71 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow 2.46=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.46\times 2}{9.81}}\\\Rightarrow t=0.71\ s

Time taken to reach the ground from the maximum height is 0.71 seconds

b) Time it stayed in the air is 0.71+0.71 = 1.42 seconds

3 0
3 years ago
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