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andrew-mc [135]
2 years ago
15

What is the density of a iphone with a mass of 200g and a volume of 40cm3

Physics
1 answer:
Leviafan [203]2 years ago
8 0
Answer: 5 gm/cc

Explanation:

200 gm/40 cc
= 5 gm/cc
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The temperature at a point (x, y) on a flat metal plate is given by T(x, y) = 54/(7 + x2 + y2), where T is measured in °C and x,
ANTONII [103]

Answer:

dt/dx = -0.373702

dt/dy =  -1.121107

Explanation:

Given data

T(x, y) = 54/(7 + x² + y²)

to find out

rate of change of temperature with respect to distance

solution

we know function

T(x, y) = 54 /( 7 + x² + y²)

so derivative it x and y direction i.e

dt/dx = -54× 2x / (7 +x² + y²)²    .........................1

dt/dy = -54× 2y / (7 + x² + y²)²      .........................2

now put the value point (1,3) as x = 1 and y = 3 in equation 1 and 2

dt/dx = -54× 2(1) / (7 +(1)² + (3)²)²  

dt/dx = -0.373702

and

dt/dy =  -54× 2(3) / (7 + (1)² + (3)²)²

dt/dy =  -1.121107

7 0
2 years ago
Which device provides electrical energy to run an electric circuit
REY [17]

The correct answer is

C. The battery

The battery is a device that provides a potential difference in the circuit, and so an electromotive force (e.m.f.) which pushes the electrons in the circuit from the negative pole towards the positive pole of the battery, so they move through the circuit. Therefore, it provides electrical energy.

8 0
3 years ago
Read 2 more answers
Explain what happens to the pitch of a cell phone ringing when the amplitude of a sound wave increases.
rewona [7]
As the amplitude of a sound wave increases the pitch of the ringing would be much higher (like if you were to inhale helium.. just with a phone)
3 0
3 years ago
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A hockey player hits a rubber puck from one side of the rink to the other. It has a mass of .170 kg, and is hit at an initial sp
Dimas [21]

By using third law of equation of motion, the final velocity V of the rubber puck is 8.5 m/s

Given that a hockey player hits a rubber puck from one side of the rink to the other. The parameters given are:

mass m =  0.170 kg

initial speed u = 6 m/s.

Distance covered s = 61 m

To calculate how fast the puck is moving when it hits the far wall means we are to calculate final speed V

To do this, let us first calculate the kinetic energy at which the ball move.

K.E = 1/2mU^{2}

K.E = 1/2 x 0.17 x 6^{2}

K.E = 3.06 J

The work done on the ball is equal to the kinetic energy. That is,

W = K.E

But work done = Force x distance

F x S = K.E

F x 61 = 3.06

F = 3.06/61

F = 0.05 N

From here, we can calculate the acceleration of the ball from Newton second law

F = ma

0.05 = 0.17a

a = 0.05/0.17

a = 0.3 m/s^{2}

To calculate the final velocity, let us use third equation of motion.

V^{2} = U^{2} + 2as

V^{2}  = 6^{2} + 2 x 0.3 x 61

V^{2} = 36 + 36

V^{2} = 72

V = \sqrt{72}

V = 8.485 m/s

Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.

Learn more about dynamics here: brainly.com/question/402617

5 0
2 years ago
He graph below shows the velocity f(t) of a runner during a certain time interval:
Annette [7]
<span>The following that describes the intercepts on the graph is "The initial velocity of the runner was 4 m/s, and the runner stopped after 8 seconds." It is because the starting point of the line is at 4 and then the ending point is at 8.

</span>
5 0
2 years ago
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