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Nady [450]
3 years ago
8

The question is in the picture

Physics
1 answer:
Sedbober [7]3 years ago
7 0

Answer:

e) 120m/s

Explanation:

When the ball reaches its highest point, its velocity becomes zero, meaning

v_0-gt = 0.

where v_0 is the initial velocity.

Solving for t we get

t = \dfrac{v_0}{g}

which is the time it takes the ball to reach the highest point.

Now, after the ball has reached its highest point, it turns around and falls downwards. After time t_0 since it had reached the highest point, the ball has traveled downwards and the velocity v_f it has gained is

v_f = gt_0,

and we are told that this is twice the initial velocity v_0; therefore,

v_f = 2v_0  = gt_0

which gives

t_0 = \dfrac{2v_0}{g}.

Thus, the total time taken to reach velocity 2v_0 is

t_{tot} = t+t_0 = \dfrac{v_0}{g}+\dfrac{2v_0}{g}

t_{tot} = \dfrac{3v_0}{g}.

This t_{tot}, we are told, is 36 seconds; therefore,

36= \dfrac{3v_0}{g},

and solving for v_0 we get:

v_0 = \dfrac{36g}{3}

v_0 = \dfrac{36s(10m/s^2)}{3}

\boxed{v_0 = 120m/s}

which from the options given is choice e.

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A 0.20-kg particle moves along the x axis under the influence of a conservative force. The potential energy is given by
anastassius [24]

Answer:

e) 11 m/s

Explanation:

For a particle under the action of a conservative force, its mechanical energy at point 0 is must be equal to its mechanical energy at point 1:

K_1+U_1=K_0+U_0\\\\\frac{mv_1^2}{2}+[(8.0\frac{J}{m^2})(x_1)^2+(2.0\frac{J}{m^4})(x_1)^4]=\frac{mv_0^2}{2}+[(8.0\frac{J}{m^2})(x_0)^2+(2.0\frac{J}{m^4})(x_0)^4]

In x_1=1.0m the speed is given, so v_1=5.0\frac{m}{s} and x_0=0. Replacing:

\frac{mv_1^2}{2}+[(8.0\frac{J}{m^2})(1m)^2+(2.0\frac{J}{m^4})(1m)^4]=\frac{mv_0^2}{2}+[(8.0\frac{J}{m^2})0^2+(2.0\frac{J}{m^4})(0)^4]\\\frac{mv_1^2}{2}+8.0J+2.0J=\frac{mv_0^2}{2}\\\frac{mv_0^2}{2}=\frac{mv_1^2}{2}+10J\\v_0=\sqrt{\frac{2}{m}(\frac{mv_1^2}{2}+10J)}\\v_0=\sqrt{\frac{2}{0.2kg}(\frac{(0.2kg)(5\frac{m}{s})^2}{2}+10J)}\\v_0=11.18\frac{m}{s}

7 0
3 years ago
Can you find a vector quantity that has a magnitude of zero but components that are not zero? Explain. Can the magnitude of a ve
MariettaO [177]

It is not possible to find a vector quantity of magnitude zero but components different from zero

The magnitude can never be less than the magnitude of any of its components

4 0
3 years ago
An 8 foot metal guy wire is attached to a broken stop sign to secure its position until repairs can be made. Attached to a stake
Angelina_Jolie [31]

Answer:

Explanation:

The guy wire is making a right angled triangle  with the ground and stop sign . It makes an angle of 51 degree with the ground. In this triangle stop sign is the perpendicular and distance from the base on the ground forms the base of the triangle . Wire forms the hypotenuse.

base / hypotenuse = cos51

base = hypotenuse x cos51

= 8 x cos51

= 5.03 ft .

The distance of the stake with which guy wire was attached from the foot of the stop sign is 5.03 ft .

7 0
3 years ago
Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 2.20×106 N , one at an angle 19.0 ∘ west of north,
valina [46]

Answer:

=2.99\times10^9J

Explanation:

According to the question

net force F = 2.20×10^6 N

displacement S = 0.72\times10^3m

from figure , the horizontal forces are same in magnitude and opposite direction.

so , neglect these two forces.

we can take only vertical components of the force.

total force F' = F cos 19° + F cos 19°

= 2×F×cos 19°   ................. (1

therefore , total work is

W = F'S

= (2F cos19)×S

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=2.99\times10^9J

6 0
3 years ago
Carbon-14 is used to determine the time an organism was living. The amount of carbon-14 an organism has is constant with the atm
lutik1710 [3]

Answer:

The age of the organism is approximately 11460 years.

Explanation:

The amount of carbon-14 decays exponentially in time and is defined by the following equation:

\frac{n(t)}{n_{o}} = e^{-\frac{t}{\tau} } (1)

Where:

n_{o} - Initial amount of carbon-14.

n(t) - Current amount of carbon-14.

t - Time, measured in years.

\tau - Time constant, measured in years.

Then, we clear the time within the formula:

t = -\tau \cdot \ln \frac{n(t)}{n_{o}} (2)

In addition, time constant can be calculated by means of half-life of carbon-14 (t_{1/2}), measured in years:

\tau = \frac{t_{1/2}}{\ln 2}

If we know that \frac{n(t)}{n_{o}} = 0.25 and t_{1/2} = 5730\,yr, then the age of the organism is:

\tau = \frac{5730\,yr}{\ln 2}

\tau \approx 8266.643\,yr

t = -(8266.643\,yr)\cdot \ln 0.25

t \approx 11460.001\,yr

The age of the organism is approximately 11460 years.

8 0
3 years ago
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