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2 years ago

Can a apple fall from a tree and not break open?

2 answers:

6 0

Well as you've seen before most likely, an apple can fall right off a tree, and stay perfectly fine. its all dependent on the variables. What angle is the apple falling at? Whats the speed, height, even the size of the apple? We must take into aware anything can happen, but its all behind the variables.

Brainliest if helpful

P.S Fact Check with the second person who just answered

Mrac [35]2 years ago

3 0

Answer:

yes

Explanation:

i had an apple tree and it happens all the time most often they don't break open.

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One horsepower is equal to how many watts

Keith_Richards [23]

One horsepower is about 745.7 watts

5 0

3 years ago

Why is temperature scalar?

shepuryov [24]

Temperature is just a measure of how HOT or COLD a substance is, which can be easily defined by a magnitude using a numerical value say “300 K” or “27°C”. Hence we can say it is a scalar quantity.

But the energy which transfer by virtue of a temperature difference is a vector quantity, as it has both magnitude and direction of motion (from High temperature to low temperature region).

3 0

2 years ago

Having difficulty finding the PE and KE for these values no mass is given. Does anyone know to go solve these?

Alexandra [31]

11) $1.04\cdot 10^7 J$

12) $1.04\cdot 10^7 J$

13) 50.0 m/s

14) 41.6 m/s

Explanation:

11)

The potential energy of an object is the energy possessed by the object due to its position relative to the ground. It is given by

$PE=mgh$

where

m is the mass of the object

g is the acceleration due to gravity

h is the height relative to the ground

Here in this problem, when the train is at the top, we have:

m = 8325 kg (mass of the train + riders)

$g=9.8 m/s^2$ (acceleration due to gravity)

h = 127 m (height of the train at the top)

Substituting,

$PE=(8325)(9.8)(127)=1.04\cdot 10^7 J$

12)

According to the law of conservation of energy, the total mechanical energy of the train must be conserved (in absence of friction). So we can write:

$KE_t + PE_t = KE_b + PE_b$

where

$KE_t$ is the kinetic energy at the top

$PE_t$ is the potential energy at the top

$KE_b$ is the kinetic energy at the bottom

$PE_b$ is the potential energy at the bottom

The kinetic energy is the energy due to motion; since the train is at rest at the top, we have

$KE_t=0$

Also, at the bottom the height is zero, so the potential energy is zero

$PE_b=0$

Therefore, we find:

$KE_b=PE_t=1.04\cdot 10^7 J$

13)

The kinetic energy of an object is the energy of the object due to its motion. Mathematically, it is given by

$KE=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the speed of the object

From question 12), we know that the kinetic energy of the train at the bottom is

$KE=1.04\cdot 10^7 J$

We also know that the mass is

m = 8325 kg

Therefore, we can calculate the speed of the train at the bottom:

$v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(1.04\cdot 10^7)}{8325}}=50.0 m/s$

14)

At the top of the second hill, the total mechanical energy of the train is still conserved.

Therefore, we can write again:

$KE_1 + PE_1 = KE_2 + PE_2$

where

$KE_1$ is the kinetic energy at the top of the 1st hill

$PE_1$ is the potential energy at the top of the 1st hill

$KE_2$ is the kinetic energy at the top of the 2nd hill

$PE_2$ is the potential energy at the top of the 2nd hill

From the previous questions, we know that

$KE_1=0$

and

$PE_1=1.04\cdot 10^7 J$

The height of the second hill is

h = 39 m

So we can also find the potential energy at the second hill:

$PE_2=mgh=(8325)(9.8)(39)=3.2\cdot 10^6 J$

So, the kinetic energy at the second hill is

$KE_2=PE_1-PE_2=1.04\cdot 10^7 - 3.2\cdot 10^6 =7.2\cdot 10^6 J$

And so, the speed is

$v=\sqrt{\frac{2KE_2}{m}}=\sqrt{\frac{2(7.2\cdot 10^6)}{8325}}=41.6 m/s$

4 0

3 years ago

An athlete completes 1 laps around a track with a radius of 25 meters in 180 seconds. What is the magnitude of the athlete's tan

DanielleElmas [232]

Answer:

0.872m/s

Explanation:

Tangential velocity is given by the formula,

$v= 2\pi r/ t$

In the question given,

radius= 25meters

time= 180secs

pie= 3.14

number of laps= 1

The magnitude of tangential velocity equals;

$\frac{1lap* 2 *3.14 *25m}{180secs}$

v = 157m/180secs

Therefore, the magnitude of the tangential velocity

=0.872m/secs

5 0

2 years ago

How many electrons will constitute 2A current in unit time

soldi70 [24.7K]

Answer:

2 charges of electron (2C)

Explanation:

I = Q/t

2 = Q/1

Q = 2×1= 2C

Q = 2 charge of electron

4 0

3 years ago