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VLD [36.1K]
2 years ago
5

A heavy frog and a light frog jump straight up into the air. They push off in such away that they both have the same kinetic ene

rgy just as they leave the ground. Air resistance is negligible. Which of the following statements about these frogs are correct?
The heavier frog goes higher than the lighter frog
The heavier frog goes higher than the lighter frog
They collide and stick together.
the lighter frog is moving faster than the heavier frog.
Physics
1 answer:
Ilia_Sergeevich [38]2 years ago
5 0

Answer:

The lighter frog goes higher than the heavier frog.

The lighter frog is moving faster than the heavier frog

Explanation:

If both frogs have the same kinetic energy when they leave the ground, the following equality applies:

K(light) = K(heavy) = \frac{1}{2} *ml*vol^{2} = \frac{1}{2}*mh*voh^{2}

Now, if the only force acting on the frogs is gravity, when they reach to the maximum height, we can apply the following kinematic equation:

vf^{2} -vo^{2} = 2*a*hmax = vf^{2} -vo^{2} = 2*(-g)*hmax

When h= hmax, the object comes momentarily to an stop, so vf =0

Solving for hmax:

hmax =\frac{vo^{2} }{2*g}

As the lighter frog, in order to have the same kinetic energy than the heavier one, has a greater initial velocity, it will go higher than the other.

As a consequence of both having the same kinetic energy, the lighter frog will be moving faster than the heavier frog.

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An object moving at a velocity of 32m/s slows to a stop in 4 seconds. What was its acceleration?
Romashka [77]

Answer:

8m/s

Explanation:

a=d/t

a=32/4

a=8 m/s

6 0
2 years ago
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A team of astronauts is on a mission to land on and explore a large asteroid. In addition to collecting samples and performing e
Blizzard [7]
<h2>Answer: 117.626m/s</h2>

Explanation:

The escape velocity V_{esc} is given by the following equation:

V_{esc}=\sqrt{\frac{2GM}{R}}   (1)

Where:

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M  is the mass of the asteroid

R  is the radius of the asteroid

On the other hand, we know the density of the asteroid is \rho=3.84(10)^{8}g/m^{3} and its volume is V=2.17(10)^{12}m^{3}.

The density of a body is given by:

\rho=\frac{M}{V}  (2)

Finding M:

M=\rhoV=(3.84(10)^{8} g/m^{3})(2.17(10)^{12}m^{3})  (3)

M=8.33(10)^{20}g=8.33(10)^{17}kg  (4)  This is the mass of the spherical asteroid

In addition, we know the volume of a sphere is given by the following formula:

V=\frac{4}{3}\piR^{3}   (5)

Finding R:

R=\sqrt[3]{\frac{3V}{4\pi}}   (6)

R=\sqrt[3]{\frac{3(2.17(10)^{12}m^{3})}{4\pi}}   (7)

R=8031.38m   (8)  This is the radius of the asteroid

Now we have all the necessary elements to calculate the escape velocity from (1):

V_{esc}=\sqrt{\frac{2(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(8.33(10)^{17}kg)}{8031.38m}}   (9)

Finally:

V_{esc}=117.626m/s This is the minimum initial speed the rocks need to be thrown in order for them never return back to the asteroid.

6 0
2 years ago
4. Dr. Copus is in charge of the cognition department at the University of Wisconsin-Madison. A new drug named Mem-Reen has beco
MrMuchimi

Answer:

See the answer below

Explanation:

<u>Independent variable</u>: Type of drug (Mem-Reen or placebo)

<u>Dependent variable</u>: memories

<u>Experimental group</u>: The group that was given Mem-Reen

<u>Control group</u>: The group that was given placebo

<u>Constants</u>: Food, hours of sleep, memory test procedures.

The independent variable is an input variable that produces effects on the dependent variable. As the variable is changed, it produces different effects on the dependent variable.

The dependent variable is the actual variable that is measured during an experiment. It is the main purpose of setting-up of an experiment.

The experimental group is also referred to as the treatment group while the control group is the group that does not receive treatment at all or they receive fake treatment/placebo.

Constants are unchanging variables included in experiments. They remain unchanged both in the treatment and the control group, otherwise, the outcome of the experiment will be unreliable.

5 0
2 years ago
Two blocks, with masses m and 3m, are attached to the ends of a string with negligible mass that passes over a pulley, as shown
olasank [31]

Answer:

a)  v = √ g x , b)  W = 2 m g d , c)    a = ½ g

Explanation:

a) For this exercise we use Newton's second law, suppose that the block of mass m moves up

            T-W₁ = m a

            W₃ - T = M a

            w₃ - w₁ = (m + M) a

            a = (3m - m) / (m + 3m) g

            a = 2/4 g

            a = ½ g

the speed of the blocks is

          v² = v₀² + 2 ½ g x

          v = √ g x

b) Work is a scalar, therefore an additive quantity

light block s

           W₁ = -W d = - mg d

3m heavy block

             

            W₂ = W d = 3m g d

the total work is

             W = W₁ + W₂

             W = 2 m g d

c) in the center of mass all external forces are applied, they relate it is

                      a = ½ g

8 0
2 years ago
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dalvyx [7]

Explanation:

It is given that,

Mass of the car 1, m_1=900\ kg

Initial speed of car 1, u_1=15i\ m/s (east)

Mass of the car 2, m_2=750\ kg

Initial speed of car 2, u_1=20j\ m/s (north)

(b) As the cars stick together. It is a case of inelastic collision. Let V is the common speed after the collision. Using the conservation of momentum as :

m_1u_1+m_2u_2=(m_1+m_2)V

900\times 15i +750\times 20j=(900+750)V

13500i+15000j=1650V

V=(8.18i+9.09j)\ m/s

The magnitude of speed,

|V|=\sqrt{8.18^2+9.09^2}

V = 12.22 m/s

(b) Let \theta is the direction the wreckage move just after the collision. It is given by :

tan\theta=\dfrac{v_y}{v_x}

tan\theta=\dfrac{9.09}{8.18}

\theta=48.01^{\circ}

Hence, this is the required solution.

4 0
2 years ago
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