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Leviafan [203]
3 years ago
8

“You can’t get more energy from a machine than the amount of energy you put into it,” is one way of stating the law of conservat

ion of _____. A. work B. friction C. energy D. power
Physics
1 answer:
goldenfox [79]3 years ago
5 0

Answer:

Energy

Explanation:

One ways of stating the law of conservation of energy is that "you cannot get more energy from a machine than the amount of energy you put into it".

The law of conservation of energy states that in a system, energy is neither created nor destroyed but transformed from one form to another.

The energy in a system does not increase based on the premise of this law.

  • The efficiency of a machine can reduce but the energy is always conserved.
  • This is because energy is transformed to other forms.
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During a tennis serve, a racket is given an angular acceleration of magnitude 150 rad/s^2. At the top of the serve, the racket h
QveST [7]

Answer:

270 m/s²

Explanation:

Given:

α = 150 rad/s²

ω = 12.0 rad/s

r = 1.30 m

Find:

a

The acceleration will have two components: a radial component and a tangential component.

The tangential component is:

at = αr

at = (150 rad/s²)(1.30 m)

at = 195 m/s²

The radial component is:

ar = v² / r

ar = ω² r

ar = (12.0 rad/s)² (1.30 m)

ar = 187.2 m/s²

So the magnitude of the total acceleration is:

a² = at² + ar²

a² = (195 m/s²)² + (187.2 m/s²)²

a = 270 m/s²

3 0
4 years ago
A stone with a mass m is dropped from an airplane that has a horizontal velocity v at a height h above a lake. If air resistance
seropon [69]

Answer: Option B. R = (1/2)gt^2

Explanation:

S = R (horizontal distance)

V^2 = 2gS

V^2 = 2gR

R = V^2 / 2g

But V = gt

R = (gt)^2 / 2g

R = (g^2 x t^2) / 2g

R = gt^2 / 2

But t^2 = 2h/g

R = ( g x 2h/g) / 2

R = h

But h = (1/2)gt^2

R = h = (1/2)gt^2

4 0
3 years ago
If the car’s speed decreases at a constant rate from 64 mi/h to 30 mi/h in 3.0 s, what is the magnitude of its acceleration, ass
mixas84 [53]

Answer:3.874 m/s^2

Explanation:

Given

Car speed decreases at a constant rate from 64 mi/h to 30 mi/h

in 3 sec

60mi/h \approx 26.8224m/s

34mi/h \approx 15.1994 m/s

we know acceleration is given by =\frac{velocity}{Time}

a=\frac{15.1994-26.8224}{3}

a=-3.874 m/s^2

negative indicates that it is stopping the car

Distance traveled

v^2-u^2=2as

\left ( 15.1994\right )^2-\left ( 26.8224\right )^2=2\left ( -3.874\right )s

s=\frac{488.419}{2\times 3.874}

s=63.038 m

7 0
3 years ago
Volume of an block is 5 cm3. If the density of the block is 250 g/cm3, what is the mass of the block ?​
9966 [12]

Answer:

1.25kg

Explanation:

Simply multiply volume and density together

4 0
3 years ago
An object is thrown upward with some velocity. If the object rises 77.5 m above the point of release, (a) how fast was the objec
jolli1 [7]

Answer:

v_o=39\ m/s\\t_m=4\ s

Explanation:

<u>Vertical Launch Upwards</u>

In a vertical launch upwards, an object is launched vertically up from a height H without taking into consideration any kind of friction with the air.

If vo is the initial speed and g is the acceleration of gravity, the maximum height reached by the object is given by:

\displaystyle h_m=H+\frac{v_o^2}{2g}

The object referred to in the question is thrown from a height H=0 and the maximum height is hm=77.5 m.

(a)

To find the initial speed we solve for vo:

\displaystyle v_o=\sqrt{2gh_m}

v_o=\sqrt{2\cdot 9.8\cdot 77.5}

v_o=39\ m/s

(b)

The maximum time or the time taken by the object to reach its highest  point is calculated as follows:

\displaystyle t_m=\frac{v_o}{g}

\displaystyle t_m=\frac{39}{9.8}

t_m=4\ s

7 0
3 years ago
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