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Ludmilka [50]
3 years ago
12

You throw an 18.0 N rock into the air from ground level and observe that, when it is 15.0 m high, it is traveling upward at 17.0

m/s. Part A Use the work-energy principle to find the rock's speed just as it left the ground. Express your answer in meters per second. vground
Physics
1 answer:
stellarik [79]3 years ago
4 0

Answer:

Vi = 24.14 m/s

Explanation:

If we apply Law of Conservation of Energy or Work-Energy Principle here, we get: (neglecting friction)

Loss in K.E of the Rock = Gain in P.E of the Rock

(1/2)(m)(Vi² - Vf²) = mgh

Vi² - Vf² = 2gh

Vi² = Vf² + 2gh

Vi = √(Vf² + 2gh)

where,

Vi = Rock's Speed as it left the ground = ?

Vf = Final Speed = 17 m/s

g = 9.8 m/s²

h = height of rock = 15 m

Therefore,

Vi = √[(17 m/s)² + 2(9.8 m/s²)(15 m)]

Vi = √583 m²/s²

<u>Vi = 24.14 m/s</u>

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An object moving 20 m/s
yan [13]

Answer:

<u>We are given:</u>

initial velocity (u) = 20m/s

acceleration (a) = 4 m/s²

time (t) = 8 seconds

displacement (s) = s m

<u />

<u>Solving for Displacement:</u>

From the seconds equation of motion:

s = ut + 1/2 * at²

replacing the variables

s = 20(8) + 1/2 * (4)*(8)*(8)

s = 160 + 128

s = 288 m

7 0
3 years ago
Instantaneous speed is measured
VMariaS [17]

Answer:

C. At a particular instant

Explanation:

Speed is the defined as the ratio between the distance covered by an object and the time taken:

v=\frac{d}{t}

where d is the distance and t the time.

However, there are two possible measurements of speed:

- Average speed: this is the speed measured over a non-zero time interval (for example: a car moving 100 metres in 5 seconds; its average speed is

v=\frac{100 m}{5 s}=20 m/s

- Instantaneous speed: this is the speed of an object measured at a particular instant in time, so for a time interval that tends to zero. So, in the previous example, the average speed is 20 m/s but the instantaneous speed of the car at various instants of time can be different from that value.

7 0
3 years ago
A ball, which has a mass of 1.25 kg, is thrown straight up from the top of a building 225 meters tall with a velocity of 52.0 m/
Anastasy [175]

First we will find the speed of the ball just before it will hit the floor

so in order to find the speed of the cart we will first use energy conservation

KE_i + PE_i = KE_f + PE_f

\frac{1}{2}mv_i^2 + mgh = \frac{1}{2}mv_f^2 + 0

\frac{1}{2}(1.25)(52)^2 + 1.25(9.8)(225) = \frac{1}{2}(1.25)v_f^2

So by solving above equation we will have

v_f = 84.3 m/s

now in order to find the momentum we can use

P = mv

P = 1.25 \times 84.3

P = 105.4 kg m/s

3 0
3 years ago
Two friends, Al and Jo, have a combined mass of 195 kg. At the ice skating rink, they stand close together on skates, at rest an
ddd [48]

Answer:

Al's mass is 102.92  kg  

Explanation:

As there are no external forces in the horizontal direction, the horizontal net force must be zero:

F_{net} = 0

As the force is the derivative in time of the momentum, this means that the horizontal momentum is constant:

F_{net} = \frac{dp_{horizontal}}{dt} = 0

p_{horizontal_i }= p_{horizontal_f}

where the suffix i and f means initial and final respectively.

The initial momentum will be:

p_{horizontal_}i = m_{Al} \ v_{Al_i} + m_{Jo} \ v_{Jo_i}

But, as they are at rest, initially

p_{horizontal_i} = m_{Al} * 0 + m_{Jo} * 0

p_{horizontal_i} = 0

So, this means:

p_{horizontal_f} = m_{Al} \ v_{Al_f} + m_{Jo} \ v_{Jo_f} = 0

We know that the have an combined mass of 195 kg:

m_{total} = m_{Al} + m_{Jo} = 195 \ kg.

so:

m_{Jo} = 195 \ kg - m_{Al}.

m_{Al} \ v_{Al_f} + (195 \ kg - m_{Al}) \ v_{Jo_f} = 0

m_{Al} \  v_{Al_f} - m_{Al} \  v_{Jo_f}= - 195 \ kg \  v_{Jo_f}

m_{Al} \ (v_{Al_f} - v_{Jo_f})= - 195 \ kg \ v_{Jo_f}

m_{Al} = \frac{ - 195 \ kg \ v_{Jo_f} } {  v_{Al_f} - v_{Jo_f} }

m_{Al} = \frac{195 \ kg  \ v_{Jo_f} } {    v_{Jo_f} - v_{Al_f} }

Now, we can use the values:

v_{Al_f}= 10.2 \frac{m}{s}

v_{Jo_f}= - 11.4 \frac{m}{s}

where the minus sign appears as they are moving at opposite directions

m_{Al} = \frac{195 \ kg  ( - 11.4 \frac{m}{s} ) } {   (- 11.4 \frac{m}{s}) - 10.2 \frac{m}{s} }

m_{Al} = 102.92 \ kg

and this is the Al's mass.

5 0
3 years ago
A mountaintop is a height y above the level ground. A woman measures the angle of elevation of the mountaintop to be θ when she
valentina_108 [34]
Let t = Theta and p = Phi
Tan t = y/x    Then x =y/Tant.
Tant = y/(x-d)  x-d = y/Tanp
y/Tant  - d = y/Tanp
y -d*Tanr = y*Tant/Tanp
y-y*Tant/Tanp  =  d*Tanr
y(1 - Tanr/Tanp = d*Tant

y = d*Tant/(1-Tant/Tanp)
7 0
3 years ago
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