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2 years ago

Help me pls its for my science class i need to show my work

Physics

1 answer:

Lostsunrise [7]2 years ago

6 0

Answer:

P = 5880 J

Explanation:

Given that,

The mass of the block, m = 30 kg

The block is sitting at a height of 20 m.

The block will have gravitational potential energy. The formula for gravitational potential energy is given by :

$P=mgh\\\\=30\times 9.8\times 20\\\\=5880\ J$

So, the required potential energy is equal to 5880 J.

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What are the units for the spring constant, k?

vitfil [10]

Answer:

C. newtons/meter

Explanation:

To find the units of spring constant you can use the hook's law, which says,

F = k × x where

F = force (unit Newton)

k = spring constant

x = expansion or contraction ( unit m)

So by this equation you can get that

k = F / X

So you get the units as N/m

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3 years ago

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A wire with a current of 3.40 A is to be formed into a circular loop of one turn. If the required value of the magnetic field at

Sloan [31]

Answer:

0.107 m

Explanation:

The magnetic field at the center of a current-carrying loop is given by

$B=\frac{\mu_0 I}{2r}$

where

$\mu_0$ is the vacuum permeability

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r is the radius of the loop

In this problem we have

I = 3.40 A is the current in the loop

$B=20 \mu T=20\cdot 10^{-6}T$ is the magnetic field at the centre of the loop

So, solving the formula for r we find

$r=\frac{\mu_0 I}{2B}=\frac{(12.56\cdot 10^{-7} H/m)(3.40 A)}{2(20\cdot 10^{-6} T)}=0.107 m$

7 0

2 years ago

A satellite orbits Earth. The only force on the satellite is the gravitational force exerted by Earth. How does the satellite’s

bulgar [2K]

Answer:

here we can say that acceleration of the satellite is same as the gravitational field due to Earth at that location

Explanation:

As we know that gravitational field is defined as the force experienced by the satellite per unit of mass

so we will have

$E = \frac{F}{m}$

now in order to find the acceleration of the satellite we know by Newton's II law

$F = ma$

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

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3 years ago

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If this decay has a half-life of 10.2 years, what mass of 60.8g carbon-14 will remain after 20.4 years

OLEGan [10]

20.4 years is 20.4/10.2 = 2 half-life cycles, which means a quarter of the starting mass or 15.2 g will remain after this time.

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3 years ago

A ship maneuvers to within 2.46×10³ m of an island’s 1.80 × 10³ m high mountain peak and fires a projectile at an enemy ship 6.1

Nesterboy [21]

Answer:

The distance close to the peak is 597.4 m.

Explanation:

Given that,

Distance of the first ship from the mountain $d=2.46\times10^{3}\ m$

Height of island $h=1.80\times10^{3}\ m$

Distance of the enemy ship from the mountain $d'=6.10\times10^{2}\ m$

Initial velocity $v=2.55\times10^{2}\ m/s$

Angle = 74.9°

We need to calculate the horizontal component of initial velocity

Using formula of horizontal component

$v_{x}=v\cos\theta$

Put the value into the formula

$v_{x}=2.55\times10^{2}\cos74.9$

$v_{x}=66.42\ m/s$

We need to calculate the vertical component of initial velocity

Using formula of vertical component

$v_{y}=v\sin\theta$

Put the value into the formula

$v_{y}=2.55\times10^{2}\sin74.9$

$v_{y}=246.19\ m/s$

We need to calculate the time

Using formula of time

$t=\dfrac{d}{v_{x}}$

$t=\dfrac{2.46\times10^{3}}{66.42}$

$t=37.03\ sec$

We need to calculate the height of the shell on reaching the mountain

Using equation of motion

$H= v_{y}t-\dfrac{1}{2}gt^2$

Put the value in the equation

$H=246.19\times37.03-\dfrac{1}{2}\times9.8\times(37.03)^2$

$H=2397.4\ m$

We need to calculate the distance close to the peak

Using formula of distance

$H'=H-h$

Put the value into the formula

$H'=2397.4-1800$

$H'=597.4\ m$

Hence, The distance close to the peak is 597.4 m.

6 0

3 years ago