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aniked [119]
3 years ago
10

Alessandra jumps on a trampoline. It takes her 1 second to

Physics
1 answer:
svp [43]3 years ago
5 0

Answer:

The speed of Alessandra's motion is 2 ft/s up.

(C) is correct option.

Explanation:

Given that,

Time = 1 sec

Height = 2 feet

We need to calculate the speed of Alessandra's motion

Using formula of speed

v=\dfrac{d}{t}

Where, v = speed

d = height

t = time

Put the value into the formula

v=\dfrac{2}{1}

v=2\ ft/s

The direction will be up ward.

Hence, The speed of Alessandra's motion is 2 ft/s up.

(C) is correct option.

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The greater the mass of an object, the greater its force due to gravity
Serga [27]
What is the question?
6 0
3 years ago
A sample of a gas has a volume of 639 cm3 when the pressure is 75.9 kPa. What is the volume of the gas when the pressure is incr
const2013 [10]

Answer:

388 cm^3

Explanation:

For this problem, we can use Boyle's law, which states that for a gas at constant temperature, the product between pressure and volume remains constant:

pV=const.

which can also be rewritten as

p_1 V_1 = p_2 V_2

In our case, we have:

p_1 = 75.9 kPa is the initial pressure

V_1 = 639 cm^3 is the initial volume

p_2 = 125 kPa is the final pressure

Solving for V2, we find the final volume:

v_2 = \frac{p_1 V_1}{p_2}=\frac{(75.9)(639)}{125}=388 cm^3

7 0
3 years ago
One end of a metal rod is in contact with a thermal reservoir at 745. K, and the other end is in contact with a thermal reservoi
Masteriza [31]

Answer:

a)S_1=-9.65}\ J/K

b)S_2=71.18\ J/K

c) 0 J/K

d)S= 61.53 J/K

Explanation:

Given that

T₁ = 745 K

T₂ = 101 K

Q= 7190 J

a)

The entropy change of reservoir 745 K

S_1=-\dfrac{7190}{745}\ J/K

Negative sign because heat is leaving.

S_1=-9.65}\ J/K

b)

The entropy change of reservoir 101 K

S_2=\dfrac{7190}{101}\ J/K

S_2=71.18\ J/K

c)

The entropy change of the rod will be zero.

d)

The entropy change of the system

S= S₁ + S₂

S = 71.18 - 9.65 J/K

S= 61.53 J/K

3 0
3 years ago
30 POINTS!!! CAN U AWNSER IT?? :)
solniwko [45]

Answer:

5235.84 kg

Explanation:

There is one theorem - whose proof I will never remember without having to drag calculus in there - that says that the variation of momentum is equal to the force applied times the time the application last.

F\Delta t = m \Delta v As long as the engine isn't ejecting mass - at this point it's a whole new can of worm - we know the force, we know the variation in speed, time to find the mass. But first, let's convert the variation of speed in meters per second. The ship gains 250 kmh, \Delta v = 69.4 m/s;

45 450 \cdot 8 = 69.4 m \rightarrow m = \frac{45450\cdot 8}{69.4} = 5235.84 kg

7 0
2 years ago
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