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3 years ago

Pick one :))) hehe please

Physics

2 answers:

Arte-miy333 [17]3 years ago

8 0

thermal energy is the correct answer of number 7

jasenka [17]3 years ago

6 0

Number 6 is D. Mantle

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A basketball is filled with air and has a small hole. Explain why the basketball would leak.

Assoli18 [71]

Because it has a small hole and air is like water, an area of higher pressure wants to move to a lower pressure area.

4 0

2 years ago

Read 2 more answers

An object is moving in a straight line along the y axis. As a function of time, its position is given by the equation y=3.0+4.8x

romanna [79]

Answer:

Explanation:

<u>Instant Velocity and Acceleration </u>

Give the position of an object as a function of time y(x), the instant velocity can be obtained by

$v(x)=y'(x)$

Where y'(x) is the first derivative of y respect to time x. The instant acceleration is given by

$a(x)=v'(x)=y''(x)$

We are given the function for y

$y(x)=3.0+4.8x +6.4x^2$

Note we have changed the last term to be quadratic, so the question has more sense.

The velocity is

$v(x)=y'(x)=4.8+12.8x$

And the acceleration is $a(x)=v'(x)=12.8$

5 0

3 years ago

A force of 150 N is applied to a 25 cm- piston in a hydraulic machine. What is the area

maxonik [38]

Answer:

8.3 cm^2

.................

6 0

3 years ago

A car travels along a highway with a velocity of 24 m/s, west. The car exits the highway; and 4.0 s later, its instantaneous vel

zloy xaker [14]

Answer:

$4.25 m/s^{2}$

Explanation:

Change in velocity considering the x component will be

Final velocity-Initial velocity

$\triangle v_x= 16cos 45^{\circ}-24=-12.6862915 m/s$

Change in velocity considering the y component will be

Final velocity-Initial velocity

$\triangle v_y= 16sin 45^{\circ}-0=11.3137085 m/s$

Resultant change in velocity $=\sqrt {(-12.6862915 m/s)^{2}+(11.3137085 m/s)^{2}}=16.9982938 m/s$

Acceleration= change in velocity per unit time hence

$a= \frac {16.9982938}{4}=4.24957345\approx 4.25 m/s^{2}$

5 0

3 years ago

A 86g ball is dropped vertically to the floor from a height of 2.87m and bounces to a height of 1.28. What is the magnitude of t

irga5000 [103]

Answer:

The impulse received by the ball from the floor during the bounce is approximately 1.11329438 m·kg/s

Explanation:

The given mass of the ball, m = 86 g = 0.089 kg

The height from which the ball is dropped, H = 2.87 m

The height to which the ball bounces, h = 1.28 m

Mathematically, we have;

Δp = F·Δt

Where;

Δp = The change in momentum = m·Δv

F = The applied force

Δt = The time of contact with the force

The velocity of the ball just before it touches the ground, v₁ = -√(2·g·H)

The velocity with which the ball leaves, v₂ = √(2·g·h)

The change in momentum, Δp = m·(v₂ - v₁)

∴ Δp = m·(√(2·g·h) - (-√(2·g·H))) = m·(√(2·g·h) +√(2·g·H) )

The impulse, Δp, received by the ball from the floor during the bounce is given as follows;

Δp = 0.089 kg × (√(2 × 9.8 m/s² × 1.28 m) + √(2 × 9.8 m/s² × 2.87 m)) ≈ 1.11329438 m·kg/s

The impulse received by the ball from the floor during the bounce, Δp ≈ 1.11329438 m·kg/s

6 0

3 years ago