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tatiyna
2 years ago
12

A frog leaps up from the ground and lands on a step 0.1 m above the ground 2 s later. We want to find the vertical velocity of t

he frog at the moment that it left the ground. Ignore air resistance. Which kinematic formula would be used
Physics
2 answers:
zhuklara [117]2 years ago
7 0

Answer:

<em>You would use the kinematic formula:</em>

    \Delta y=V_{0y}\times t-g\times t^2/2

Explanation:

The upwards vertical motion is ruled by the equation:

        y=y_0+V_{0y}\times t-g\times t^2/2

Where:

       y \text{ is the position at the time }t:y=0.1m

       y_0\text{ is the initial position: }y_0=0

       t=2s

       g\text{ is the gravitational acceleration: }\approx 9.8m/s^2

       V_{0y}\text{ is the initial vertical velocity}

Naming Δy = y - y₀, the equation becomes:

      \Delta y=V_{0y}\times t-g\times t^2/2

Then, you just need to substitute with Δy = 0.1m, t = 2s, and g = 9.8m/s², ans solve for the intital vertical velocity.

andrezito [222]2 years ago
7 0

Answer: the equation you would use is

v = v0 +at

Explanation:

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On a cold winter day when the temperature is −20∘C, what amount of heat is needed to warm to body temperature (37 ∘C) the 0.50 L
vlabodo [156]

Answer:

75.6J

Explanation:

Hi!

To solve this problem we must use the first law of thermodynamics that states that the heat required to heat the air is the difference between the energy levels of the air when it enters and when it leaves the body,

Given the above we have the following equation.

Q=(m)(h2)-(m)(h1)

where

m=mass=1.3×10−3kg.

h2= entalpy at 37C

h1= entalpy at -20C

Q=m(h2-h1)

remember that the enthalpy differences for the air can approximate the specific heat multiplied by the temperature difference

Q=mCp(T2-T1)

Cp= specific heat of air = 1020 J/kg⋅K

Q=(1.3×10−3)(1020)(37-(-20))=75.6J

4 0
2 years ago
An object with a heat capacity of 345J∘C experiences a temperature change from 88.0∘C to 45.0∘C. How much heat is released in th
pogonyaev

Answer:

There is 148.35 Joules of heat is released in the process.

Explanation:

Given that,

Heat capacity of the object, c=345J/^oC

Initial temperature, T_i=88^{\circ}C

Final temperature, T_f=45^{\circ}C

We need to find the amount of heat released in the process. It is a concept of heat capacity. The heat released in the process is given by :

Q=mc\Delta T

Let the mass of the object is 10 g or 0.01 kg

So,

Q=0.01\times 345\times (88-45)

Q = 148.35 Joules

So, there is 148.35 Joules of heat is released in the process. Hence, this is the required solution.                                                

5 0
2 years ago
The Millersburg Ferry (m = 13000.0 kg loaded) is travelling at 11 m/s when the engines are put in reverse. The engineproduces a
Pie

Explanation:

It is given that,

Mass of Millersburg Ferry, m = 13000 kg

Velocity, v = 11 m/s

Applied force, F = 10⁶ N

Time period, t = 20 seconds

(a) Impulse is given by the product of force and time taken i.e.

J=F.\Delta t

J=10^6\ N\times 20\ s

J=2\times 10^7\ N-s

(b) Impulse is also given by the change in momentum i.e.

J=\Delta p=p_f-p_i

J=p_f-p_i

p_f=J+p_i

p_f=2\times 10^7\ N-s+13000\ kg\times 11\ m/s

p_f=20143000\ kg-m/s

(c) For new velocity,

v_f=\dfrac{p_f}{m}

v_f=\dfrac{20143000\ kg-m/s}{13000\ kg}

v_f=1549.46\ m/s

Hence, this is the required solution.

3 0
2 years ago
Read 2 more answers
(a) A submarine descends to a depth of 70 m below the surface of water. The density of the water is 1050 kg/m3. Atmospheric pres
timurjin [86]

Answer:

sttouyietETwe2e664yrwtwwteuwtrwruwuuwwuwtwuw7w7w5w7w772253536464647

5 0
1 year ago
You run due east at a constant speed of 3.00 m/s for a distance of 120.0 m and then continue running east at a constant speed of
Leni [432]

Answer:

Explanation:

Given

Speed while running towards east is v_1=3\ m/s

Distance traveled in east direction x_1=120\ m

For Another interval you  run with velocity

v_2=5\ m/s

x_2=240\ m

Total displacement=x_1+x_2

=120+120=240\ m

Time for first interval

t_1=\frac{x_1}{v_1}=\frac{120}{3}

t_1=\frac{120}{3}=40\ s

Time for second interval

t_2=\frac{x_2}{v_2}=\frac{120}{5}=24\ s

total time t=t_1+t_2

t=40+24=64\ s

average velocity v_{avg}=\frac{x_1+x_2}{t}

v_{avg}=\frac{240}{64}=3.75\ m/s

Therefore average velocity is less than 4 m/s  

7 0
2 years ago
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