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3 years ago

I need helpppppppppppppppppp!

Physics

2 answers:

allsm [11]3 years ago

3 0

Answer:

i think its B

Explanation:

but check it before do it

Sedbober [7]3 years ago

3 0

Answer:

Explanation:

Hope it helps I tried looking for think I found it hit the heart pls

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What is the forms of energy given below.

Damm [24]

Answer:

Lightening of the table lamp

Explanation:

Energy has a different form of energy. In physics, the capacity of the form of energy is work. The energy can exist in the form of thermal, potential, kinetic, chemical and electrical, and nuclear. There are other forms of energy such as work and heat.

The energy is designated according to the nature of the objects. So that when heat transferred it has been changed into thermal. All the forms of energy are related to the motion of an object. Energy can neither destroyed or created.

3 0

3 years ago

If an astronaut can throw a certain wrench 15.0 m vertically upward on earth, how high could he throw it on our moon if he gives

Sever21 [200]

Answer:

hm= 18.2 m

Explanation:

Solution is attached

5 0

3 years ago

An astronaut holds a rock 100 m above surface of Planet X. The rock is then thrown upwards with a sleek of 15m/s. The rock reach

Gelneren [198K]

Answer: $5 m/s^{2}$

Explanation:

This problem is related to vertical motion, and the equation that models it is:

$y=y_{o}+V_{o}sin\theta t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0m$ is the rock's final height

$y_{o}=100 m$ is the rock's initial height

$V_{o}=15 m/s$ is the rock's initial velocity

$\theta=90\°$ is the angle at which the rock was thrown (directly upwards)

$t=10 s$ is the time

$g$ is the acceleration due gravity in Planet X

Isolating $g$ and taking into account $sin(90\°)=1$ :

$g=(-\frac{2}{t^{2}})(y-y_{o}-V_{o}t)$ (2)

$g=(-\frac{2}{(10 s)^{2}})(0 m-100 m-(15 m/s)(10 s))$ (3)

$g=5 m/s^{2}$ (4) This is the acceleration due gravity in Planet X

5 0

3 years ago

A charge of 50 µC is placed on the y axis at y = 3.0 cm and a 77-µC charge is placed on the x axis at x = 4.0 cm. If both charge

zheka24 [161]

Answer:

The acceleration of an electron is $1.2\times10^{20}\ m/s^2$

Explanation:

Given that,

One Charge = 50 μC

Distance on y axis = 3.0 cm

Second charge = 77 μC

Distance on x axis = 4.0 cm

We need to calculate the force on electron due to q₁

Using formula of force

$F_{1}=\dfrac{kq_{1}q}{r^2}$

Here, q = charge of electron

Put the value into the formula

$F_{1}=\dfrac{9\times10^{9}\times50\times10^{-6}\times1.6\times10^{-19}}{(3\times10^{-2})^2}$

$F_{1}=8\times10^{-11}j\ \ N$

We need to calculate the force on electron due to q₂

Using formula of force

$F_{2}=\dfrac{kq_{2}q}{r^2}$

Here, q = charge of electron

Put the value into the formula

$F_{2}=\dfrac{9\times10^{9}\times77\times10^{-6}\times1.6\times10^{-19}}{(4\times10^{-2})^2}$

$F_{2}=6.93\times10^{-11}i\ \ N$

We need to calculate the net force

Using formula of net force

$F=F_{1}+F_{2}$

Put the value into the formula

$F=8\times10^{-11}j+6.93\times10^{-11}i$

The magnitude of the net force

$F=\sqrt{(8\times10^{-11})^2+(6.93\times10^{-11})^2}$

$F=1.058\times10^{-10}\ N$

We need to calculate the acceleration of an electron

Using formula of force

$F = ma$

$a=\dfrac{F}{m}$

Put the value into the formula

$a=\dfrac{1.058\times10^{-10}}{9.1\times10^{-31}}$

$a=1.2\times10^{20}\ m/s^2$

Hence, The acceleration of an electron is $1.2\times10^{20}\ m/s^2$

3 0

3 years ago

For a conclusion to be accepted by the scientific community, the data must be accurate, free from error, and

Setler79 [48]

The data must be reproducible: it means that the data will be reliable and representative and not that this data would lead to one conclusion and another data would lead to another.

Data being varied, unique or surprising does not have anything to do with acceptability of the conclusion.

8 0

3 years ago

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