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morpeh [17]
3 years ago
5

Show with proof, the planet which an astronaut would have a greater weight if Planet A has twice the mass and twice the radius o

f Planet B.
Physics
1 answer:
meriva3 years ago
4 0

Answer:

First let's look at the gravitational formula by newton:

f = g \times  \frac{m1 \times m2}{ {r}^{2} }

Now the f and g should be capital but that's not possible with the equation system. But we can see that the force the astronaut is pulled at is dependent on the mass and the distance if we assume his mass stays the same (mass and weight aren't the same also g is a constant). If a planet has twice the radius the force will by four times as weak because of the ^2. This is not compensated by the twice as big mass. Therefore the astronaut will have a higher force and thus a higher weight on planet B

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Un acróbata de 60.0 kg está unido a un cordón de bungee con un resorte de 10.0 m de longitud . Salta de un puente que abarca un
Ede4ka [16]

Answer:

What kind of languege is this please?

8 0
3 years ago
A platinum sphere with radius 0.0139 m is totally immersed in mercury. Find the weight of the sphere, the buoyant force acting o
Inessa [10]

Answer:

A) W = 0.59 N

B) Buoyant Force = 0.37N

C) Apparent Weight = 0.22N

Explanation:

Volume of a sphere is givwn by;

V= (4/3)πr³

We are given radius r = 0.0139 m

Thus, V = (4/3)π(0.0139³)

V = 2.81237 x 10^(-6) m³

A) Weight of sphere, W= mg

where mass, m can be expressed as; m = Volume x density of platinum

m = 2.81237 x 10^(-6) x 2.14 × 10⁴ = 6.0185 x 10^(-2) kg

So, W = 6.0185 x 10^(-2) x 9.8

W = 0.59 N

B) Buoyant force = mg

where mass of displaced mercury m can be expressed as; m = Volume x density of mercury

m = 2.81237 x 10^(-6) x 1.36 × 10⁴

m = 3.825 x 10^(-2) kg

Thus, Buoyant force = 3.825 x 10^(-2) x 9.8 = 0.37N

C) Apparent weight = Weight of sphere - Buoyant force

Thus, apparent weight = 0.59 - 0.37 = 0.22N

8 0
3 years ago
Read 2 more answers
A plane is moving due north, directly towards its destination. Its airspeed is 200 mph. A constant breeze is blowing from west t
Vikki [24]

Answer:

200 m/s

Explanation:

Given that,

A plane is moving due north, directly towards its destination. Its airspeed is 200 mph. A constant breeze is blowing from west to east at 60 mph.

We need to find the rate at which the plane is moving towards North. It can be given by :

V_x=v\cos\theta

Where

\theta is the angle with the North

V_x=200\times \cos(\dfrac{60}{200})\\\\=199.99\ m/s\\\\\approx 200\ m/s

Hence, the plane is moving at a rate of 200 m/s.

4 0
3 years ago
Electromagnetic waves are used extensively in modern technology. Many devices are built to emit and/or receive EM waves at a ver
il63 [147K]

Answer: Option D and E

Explanation: Across the electromagnet spectrum, the frequency increases but the wavelength reduces.

This implies that radio waves have the longest wavelength but the smallest frequency, while gamma rays have the shortest wavelength but the highest frequency.

This implies that there is an inverse relationship between wavelength and frequency.

Of all the technologies listed above, wireless internet has the highest frequency hence making it have the shortest wavelength.

Also by comparing values from the data given to us above, it is possible for some wireless internet device to operate within the same frequency range of the microwave.

3 0
3 years ago
An electric dipole consisting of charges of magnitude 2.00 nC separated by 8.40 μm is in an electric field of strength 1390 N/C.
amid [387]

Answer:

(a) The magnitude of the electric dipole moment is 1.68 x 10⁻¹⁴ C.m

(b) The difference between the potential energies ΔU, is 4.6704 x 10⁻¹¹ J

Explanation:

Given;

magnitude of charge, q = 2 nC = 2 x 10⁻⁹ C

distance of separation, d = 8.4 μm = 8.4 x 10⁻⁶ m

strength of electric field, E = 1390 N/C

(a) the magnitude of the electric dipole moment

p = qd

p = (2 x 10⁻⁹ C)(8.4 x 10⁻⁶ m)

p = 1.68 x 10⁻¹⁴ C.m

(b) the difference between the potential energies for dipole orientations parallel and anti-parallel to E

ΔU = U(180) - U(0)

ΔU = 2pE

ΔU = 2(1.68 x 10⁻¹⁴ )(1390)

ΔU = 4.6704 x 10⁻¹¹ J

6 0
3 years ago
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