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2 years ago

A shell is fired with 500km/h on a target half kilometer away.In what time in which shell will hit the target.

Physics

1 answer:

Drupady [299]2 years ago

6 0

It depends on the direction in which the shell is launched. The time can be anything from 3.6 seconds to never.

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The acceleration due to gravity on Jupiter is 2.5 times what it is here on earth. An object weighing 347.9 N here on earth will

inessss [21]

Answer: weight on Jupiter = 869.75 N

mass on Earth = mass on Jupiter = 35.5 Kg

Explanation:

W = mg

W = weight

m = mass

g = gravitational acceleration [ on the Earth, g₁ = 9,8 N/kg ]

On the Earth,

G₁ = m x g₁ = 347,9 N

On the Jupiter,

G₂ = mg₂

mass on the Earth = mass on the Jupiter !

m = G₁ : g = 347.9 N : 9,8 N/kg = 35.5 kg

G2 : G1 = 2.5

G₂ = 2,5 G₁ = 2,5 x 347.9 N = 869,75 N

5 0

2 years ago

Beginning at the NW corner of the intersection of Pine & 675, thence north 950 feet, thence west 380 feet, thence south 950

Serjik [45]

Answer:

this description is valid for mediadle displacement, bone is an acceptable description

Explanation:

The description of a person's position must be done with a position vector. These vectors must have magnitude, a given direction and a starting point.

In the description this has a starting point corner NO of pine and 675.

Each displacement occurs with respect to the previous one, indicating the magnitude of the displacement and its direction.

After analyzing this description is valid for mediadle displacement, bone is an acceptable description

6 0

2 years ago

How many protons are in this atom if it has a balanced charge

Lerok [7]

It would be 20 protons left because 1-19= 20

7 0

1 year ago

(a) If a proton with a kinetic energy of 6.2 MeV is traveling in a particle accelerator in a circular orbit with a radius of 0.5

Tju [1.3M]

Answer:

The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

Explanation:

Suppose Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$

Given that,

Kinetic energy = 6.2 MeV

Radius = 0.500 m

We need to calculate the acceleration

Using formula of acceleration

$a=\dfrac{v^2}{r}$

Put the value into the formula

$a=\dfrac{\dfrac{1}{2}mv^2}{\dfrac{1}{2}mr}$

Put the value into the formula

$a=\dfrac{6.2\times10^{6}\times1.6\times10^{-19}}{\dfrac{1}{2}\times1.67\times10^{-27}\times0.51}$

$a=2.32\times10^{15}\ m/s^2$

We need to calculate the rate at which it emits energy because of its acceleration is

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$

Put the value into the formula

$\dfrac{dE}{dt}=\dfrac{(1.6\times10^{-19})^2\times(2.3\times10^{15})^2}{6\pi\times8.85\times10^{-12}\times(3\times10^{8})^3}$

$\dfrac{dE}{dt}=3.00\times10^{-23}\ J/s$

The energy in ev/s

$\dfrac{dE}{dt}=\dfrac{3.00\times10^{-23}}{1.6\times10^{-19}}\ J/s$

$\dfrac{dE}{dt}=1.875\times10^{-4}\ ev/s$

We need to calculate the fraction of its energy that it radiates every second

$\dfrac{\dfrac{dE}{dt}}{E}=\dfrac{1.875\times10^{-4}}{6.2\times10^{6}}$

$\dfrac{\dfrac{dE}{dt}}{E}=3.02\times10^{-11}$

Hence, The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

5 0

2 years ago

If you are an astronaut in the middle of the near side of the moon during a full moon,how would the ground around you look?How w

Radda [10]

The moon would be bright and the earth would be darker because the sun is on the opposite side of the earth at that time and the light from it is reflecting off the moon to produce light upon the nigh also.......

You wouldn’t see the sun a night...

Unless you lived in the north/south pole

3 0

2 years ago

Read 2 more answers