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Maslowich
2 years ago
14

A small child gives a plastic frog a big push at the bottom of a slippery 2.0 meter long, 1.0 meter high ramp, starting it with

a speed of 5.0 m/s. What is the frog's speed as it flies off the top of the ramp?

Physics
1 answer:
valentinak56 [21]2 years ago
7 0
Refer to the diagram shown below.

Because the ramp is slippery, ignore dynamic friction.
Let m =  the mass of the frog.
g = 9.8 m/s²

The KE (kinetic energy) at the bottom of the ramp is
KE₁ = (1/2)*(m kg)*(5 m/s)² = 12.5 m J

Let v =  the velocity at the top of the ramp.
The KE at the top of the ramp is
KE₂ = (1/2)*m*v²= 0.5 mv² J
The PE (potential energy) at the top of the ramp relative to the bottom is
PE₂ = (m kg)*(9.8 m/s²)*(1 m) = 9.8m J

Conservation of energy requires that
KE₁ = KE₂ + PE₂
12.5m = 0.5mv² + 9.8m
0.5v² = 2.7
v = 2.324 m/s

Answer: 2.324 m/s

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A typical jet airliner has a cruise airspeed of 900 km/h , which is its speed relative to the air through which it is flying. If
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Explanation:

It is given that,

Speed of the jet airplane with respect to air, v_{PA} = 900\ km/h          

If the wind at the airliner’s cruise altitude is blowing at 100 km/h from west to east, v_{AG}= 100\ km/h

(A) Let v_{PG} is the speed of the airliner relative to the ground if the airplane is flying from west to east,

v_{PG}=900-100=800\ km/h

(B) Let v'_{PG} is the speed of the airliner relative to the ground if the airplane is flying from east to west,

v'_{PG}=900+100=1000\ km/h

Hence, this is the required solution.                                            

8 0
2 years ago
Determine the length of a copper wire that has a resistance of 0.172 ? and cross-sectional area of 7.85 × 10-5 m2. The resistivi
KonstantinChe [14]

Answer:

Length of copper wire, l = 785 meters

Explanation:

Given that,

Resistance of the copper wire, R = 0.172 ohms

Area of cross section, A=7.85\times 10^{-5}\ m^2

Resistivity of copper, \rho=1.72\times 10^{-8}\ \Omega-m

The resistance of a wire is given by :

R=\rho\dfrac{l}{A}

l=\dfrac{RA}{\rho}

l=\dfrac{0.172\ \Omega\times 7.85\times 10^{-5}\ m^2}{1.72\times 10^{-8}\ \Omega-m}

l = 785 meters

So, the length of the copper wire is 785 meters. Hence, this is the required solution.

8 0
2 years ago
A farm hand does 972 J of work pulling an empty hay wagon along level ground with a force of 310 N [23° below the horizontal]. T
irina1246 [14]

Answer:

Option d)

Solution:

As per the question:

Work done by farm hand, W_{FH} = 972J

Force exerted, F' = 310 N

Angle, \theta = 23^{\circ}

Now,

The component of force acting horizontally is F'cos\theta

Also, we know that the work done is the dot or scalar product of force and the displacement in the direction of the force acting on an object.

Thus

W_{FH} = \vec{F'}.\vec{d}

972 = 310\times dcos23^{\circ}

d = 3.406 m = 3.4 m

3 0
2 years ago
In two separate double slit experiments, an interference pattern is observed on a screen. In the first experiment, violet light
sashaice [31]

Answer:

\lambda_3 = 4.72*10^{-7} m

Explanation:

given data:

wavelength \lambda = 708nm = 708*10^{-9} m

using the following relation:

y = \frac{mL\lambda}{d}

according to the given information

second and third dark fringe is at same location. so

y_2 = y_3

\frac{m_2L\lambda_2}{d} = \frac{m_3L\lambda_3}{d}

m_2\lambda_2 = m_3\lambda_3

2*708*10^{-9} = 3*\lambda_3

\lambda_3 = \frac{2*708*10^{-9}}{3}

\lambda_3 = 4.72*10^{-7} m

4 0
2 years ago
A 3-cm high object is in front of a thin lens. The object distance is 4 cm and the image distance is –8 cm. (a) What is the foca
xenn [34]

Answer:

a) Focal length of the lens is 8 cm which is a convex lens

b) 6 cm

c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.

Explanation:

u = Object distance =  4 cm

v = Image distance = -8 cm

f = Focal length

Lens Equation

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}=\frac{1}{4}+\frac{1}{-8}\\\Rightarrow \frac{1}{f}=\frac{1}{8}\\\Rightarrow f=\frac{8}{1}=-8\ cm

a) Focal length of the lens is 8 cm which is a convex lens

Magnification

m=-\frac{v}{u}\\\Rightarrow m=-\frac{-8}{4}\\\Rightarrow m=2

b) Height of image is 2×3 = 6 cm

Since magnification is positive the image upright

c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.

8 0
2 years ago
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