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KonstantinChe [14]
3 years ago
11

Why can you jump further on the Moon than on the earth?

Physics
2 answers:
tekilochka [14]3 years ago
5 0
You do jump further because you weigh far less on the moon than earth
kicyunya [14]3 years ago
3 0
Because the gravitational field strength on the moon (1.6N/kg)is smaller than the gravitational field strength on the earth (9.8N/kg). Weight= mass×gravitational field strength. Your mass is the same but as g is smaller, your weight decreases. weight is the gravitational force, the downwards force towards the centre of the planet, and as it will be smaller on the moon, the force of attraction between the moon and you is smalller so when you jumb you will reach a higher point above the ground than when you jumb on the earth. Basically the moon 'pulls' you less than the earth.
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Calculate the radius of the orbit of a proton moving at 2.2x10^6 m/s in a magnetic field 0.7 T where v and B are perpendicular.
Juliette [100K]

Answer:

3.28 cm

Explanation:

To solve this problem, you need to know that a magnetic field B perpendicular to the movement of a proton that moves at a velocity v will cause a Force F experimented by the particle that is orthogonal to both the velocity and the magnetic Field. When a particle experiments a Force orthogonal to its velocity, the path it will follow will be circular. The radius of said circle can be calculated using the expression:

r = \frac{mv}{qB}

Where m is the mass of the particle, v is its velocity, q is its charge and B is the magnitude of the magnetic field.

The mass and  charge of a proton are:

m = 1.67 * 10^-27 kg

q = 1.6 * 10^-19 C

So, we get that the radius r will be:

r =  \frac{1.67 * 10^-27 kg * 2.2*10^6 m/s}{1.6 * 10^-19 C* 0.7 T} = 0.0328 m, or 3.28  cm.

8 0
3 years ago
What do the peaks represent
monitta
An exothermic reaction is a chemical reaction that releases heat. It gives net energy to its surroundings. That is, the energy needed to initiate the reaction is less than the energy released. ... Δ H = (energy used in forming product bonds) − (energy released in breaking reactant bonds)
6 0
3 years ago
A Cessna aircraft has a liftoff speed of 120 km/h What minimum constant acceleration does the aircraft require to be airborne af
Lady_Fox [76]

Answer:

<h3>2.3125m/s²</h3>

Explanation:

Using the equation of motion v² = u²+2aS

v is the final velocity = 120km/hr

120km/hr = 120 * 1000/1 * 3600 = 33.3m/s

u is the initial velocity = 0m/s

a is the acceleration

S is the distance covered = 240m

On substituting the given parameters

33.3² = 0²+2a(240)

33.3² = 480a

1110 = 480a

a = 1110/480

a = 2.3125m/s²

Hence the minimum constant acceleration that the aircraft require to be airborne after a takeoff run of 240 m is 2.3125m/s²

4 0
3 years ago
Gravel is being dumped from a conveyor belt at a rate of 10 ft3/min, and its coarseness is such that it forms a pile in the shap
AURORKA [14]

Answer: The height of the pile increasing is increasing at 0.16\frac{ft}{min}

Explanation:

Given

Rate at which gravel is being dumped , \frac{\mathrm{d} V}{\mathrm{d} t}=10\frac{ft^{3}}{min}

=>Rate of increase of volume of cone=\frac{\mathrm{d} V}{\mathrm{d} t}=10\frac{ft^{3}}{min}

If height of the cone at any instant is h then the diameter of cone is also h

Volume of cone , V=\frac{\pi r^{2}h}{3}=\frac{\pi h^{3}}{4\times 3}=\frac{\pi h^{3}}{12}

Now differentiate both sides w.r.t time(t)

\frac{\mathrm{d} V}{\mathrm{d} t}=\frac{\pi h^{2}}{4}\frac{\mathrm{d} h}{\mathrm{d} t}

Therefore at h = 9 ft

10=\frac{\pi \times 9^{2}}{4}\times \frac{\mathrm{d} h}{\mathrm{d} t}

=>\frac{\mathrm{d} h}{\mathrm{d} t}=0.16\frac{ft}{min}

Thus the height of the pile increasing is increasing at 0.16\frac{ft}{min}

6 0
3 years ago
What is a current?
suter [353]

Answer:

sorry, no

Explanation:

7 0
3 years ago
Read 2 more answers
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