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Basile [38]
2 years ago
12

An adult elephant has a mass of about 5.0 tons. An adult elephant shrew has a mass of about 50 grams. How far r from the center

of the Earth should an elephant be placed so that its weight equals that of the elephant shrew on the surface of the Earth
Physics
1 answer:
Diano4ka-milaya [45]2 years ago
4 0

Answer:

2023857702.507m

Explanation:

f=\frac{GMm}{r^{2} }

recall from newton's law of gravitation

G=gravitational constant

mshew=50g

melephant=5*10^3kg

rearth=radius of the earth 6400km or 6400000m

mearth= masss of the earth

Gm(shrew)m(earth)/r(earth)^2 = Gm(elephant)m(earth)/r^2

strike out the left hand side and right hand side variables

m(shrew)/r(earth)^2 = m(elephant)/r^2

r^2 = m(elephant).r(earth)^2 / m(shrew)       .........make r^2 the subject of the equation

r^2=(5*10^{3} *(6400000)^{2} )/.05

r^2=40960000000000

r=2023857702.507m

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Answer:

C

Explanation:

From above question we know that

A = 6.2 m

f = 1.6 rad/s

t = 3.5 s

x =?

We know that,

x = Acos(2pie ft)

Putting all values in above eq.

x = 6.2 x cos(2x3.142x1.6x3.5)

x = - 4.8

Displacement can never be negative so ignore - sign.

4 0
2 years ago
The figure shows two 1.0 kg blocks connected by a rope. a second rope hangs beneath the lower block. both ropes have a mass of 2
lutik1710 [3]
We need first to use the formula  F=m(a+g), m iis the total mass, a is the acceleration, g is gravity pulling the blocks. So the procedure will be 
<span>m=2kg(both blocks)+500g(both ropes) → m=2.5kg </span>
<span>a=3.00m/s^2 </span>
<span>g=9.8m/s^2 </span>
<span>F=m(a+g) → F=2.5kg (3.00m/s^2 + 9.8m/s^2) → F=2.5kg (12.8m/s^2) → F=32 N 
To calculate the tension at the top of rope 1 you need to use the formula </span>T=m(a+g) so it will be <span>T=m(a+g) → T=1.5kg(12.8m/s^2) → T=19.2N 
</span>We can now calculate the tension at the bottom of rope 1 using the formula: <span>T=m(a+g) → T=1.25kg(12.8m/s^2) → T=16N 
</span>Now to find the tension at the top of rope 2 we do it like this: 
<span>T=m(a+g) → T=.25kg(12.8m/s^2) → T=3.2</span>
7 0
1 year ago
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Sully is riding a snowmobile on a flat, snow-covered surface with a constant velocity of 10 meters/second. The total mass of the
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16 - 10 = 6
6 / 10 = .6
F = 280 x .6 = 168
7 0
2 years ago
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A wheel with rotational inertia 0.04 kg•m2 and radius 0.02 m is turning at the rate of 10 revolutions per second when a friction
Rus_ich [418]

Answer:

-78.96 J

Explanation:

The workdone by the torque in stopping the wheel = rotational kinetic energy change of wheel.

So W = 1/2I(ω₁² - ω₀²) where I = rotational inertia of wheel = 0.04 kgm², r = radius of wheel = 0.02 m, ω₀ = initial rotational speed = 10 rev/s × 2π = 62.83 rad/s, ω₁ = final rotational speed = 0 rad/s (since the wheel stops)

W = 1/2I(ω₁² - ω₀²) = 1/2 0.04 kgm² (0² - (62.83 rad/s)²) = -78.96 J  

7 0
2 years ago
A sample of silver (with work function Φ=4.52 eV ) is exposed to an ultraviolet light source (????=200 nm), which results in the
Crank

Answer:

1. c

2. e

3. d

Explanation:

1.

From Einstein's Photoelectric Equation, we know that:

Energy given up by photon = Work Function + K.E of Electron

hc/λ = φ + K.E

where,

h = Plank's Constant = 6.626 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of light source = 200 nm = 2 x 10⁻⁷ m

φ = (5.1 eV)(1.6 x 10⁻¹⁹ J/eV) = 8.16 x 10⁻¹⁹ J

Therefore,

(6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2 x 10⁻⁷ m) - 8.16 x 10⁻¹⁹ = K.E

K.E = (9.939 - 8.16) x 10⁻¹⁹ J

K.E = 1.778 x 10⁻¹⁹ J

The positive answer shows that electrons will be emitted. Since it is clear from the equation the the K.E of electron decreases with the increase in work function. Therefore:

c. <u>less energetic photo-electrons (on average)</u>

<u></u>

2.

The increase in light sources means an increase in the intensity of light. The no. of photons are increased, due to increase of intensity. Thus, more photons hit the metal and they eject greater no. of electrons. Therefore,

e. <u>more photo-electrons ejected</u>

<u></u>

3.

X-rays have smaller wavelength and greater energy than ultraviolet rays. Thus, the photons with greater energy will strike the metal and as a result, electrons with higher energy will be ejected.

d. <u>more energetic photo-electrons (on average)</u>

8 0
2 years ago
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