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WINSTONCH [101]
4 years ago
5

A 28 kg rock falls freely near the surface of the Earth. Approximately how much gravitational force is exerted by the Earth on t

he rock? A. 2.8 N B. 280 N C. 28 N D. 2,800 N
Physics
1 answer:
ValentinkaMS [17]4 years ago
4 0

Answer :Force is exerted by the Earth on the rock is 280 N .

Given :

Mass of rock falling falling , m = 28 kg .

Acceleration due to gravity , g=10\ m/s^2 .

We need to find the force exerted by the earth on the rock .

Let force exerted be F .

Now , we know by Newton's Law Of Motion .

F = ma     .......( 1 )

Here , m is  mass and a is acceleration in this case it is g=10\ m/s^2 .

Putting all these value in equation( 1 ) we get ,

F=28\times 10\\\\F=280\ N

Therefore, this is the required solution .

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The electric field 2.5 mm from a uniform sheet of charge is σ=800, NC. How much charge is contained in a 5.0x5.0 cm section of t
slamgirl [31]

Answer:

The charge is 2.75\times10^{-13}\ C

Explanation:

Given that,

Distance = 2.5 mm

Electric field = 800 NC

Length L=5.0\times5.0\times10^{-4}\ m

We need to calculate the linear charge density

Using formula of linear charge density

E=\dfrac{2k\lambda}{r}

\lambda=\dfrac{Er}{2k}

Put the value into the formula

\lambda=\dfrac{800\times2.5\times10^{-3}}{2\times9\times10^{9}}

\lambda=1.1\times10^{-10}\ C/m

We need to calculate the charge

Using formula of charge

Q=\lambda\timesL

Put the value into the formula

Q=1.1\times10^{-10}\times(5.0\times5.0\times10^{-4})

Q=2.75\times10^{-13}\ C

Hence, The charge is 2.75\times10^{-13}\ C

5 0
3 years ago
A person wishes to heat pot of fresh water from 20°C to 100°C in order to boil water for pasta. They calculate that their pot ho
nasty-shy [4]
Given:
Water, 2 kilograms
T1 = 20 degrees Celsius, T2 = 100 degrees Celsius.  

Required:
Heat produced  

Solution:
Q (heat) = nRT = nR(T2 = T1)
Q (heat) = 2 kilograms (4.184 kiloJoules per kilogram Celsius) (100 degrees Celsius – 20 degrees Celsius)
<u>Q (heat) = 669.42 Joules
</u>This is the amount of heat produced in boiling 2 kg of water.
6 0
4 years ago
Determine the kinetic energy of a 10 Kg roller coaster car that is moving with a speed of 2m/s.
skelet666 [1.2K]

In this question all required information's are already provided. Based on these details the answer to the question can be easily determined. Let us now write down all the information's that are already given.

Mass of the roller coaster = 1000 kg

Velocity of the roller coaster = 20.0 m/s

We know the formula for finding the kinetic energy is

Kinetic energy = 0.5 * mass * (velocity) ^2

                      = 0.5 * 1000 * (20)^2

                      = 0.5 * 1000 * 400

                      = 200000 Joules

So the Kinetic energy of the roller coaster is 200000 joules.

i hope this helps you friend good luck on your quiz or lesson

6 0
2 years ago
A flat, circular, steel loop of radius 75 cm is at rest in a uniform magnetic field, as shown in an edge-on view in the figure (
SIZIF [17.4K]

The solution to the questions are given as

  • t=40.39 \mathrm{sec}
  • \varepsilon &=(0.12v)e^{0.057t}
  • the direction of induced current will be Counterclock vise.

<h3>What is the direction of the current induced in the loop, as viewed from above the loop.?</h3>

Given, $B(t)=(1.4 T) e^{-0.057 t}$

$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}

\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$

\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}

\varepsilon &=(0.12v)e^{0.057t}

(b) Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$

\begin{aligned}&\therefore 1 . \varepsilon_{0}=\varepsilon_{0} \cdot e^{-e .057 t} \\&\therefore e^{0.057 t}=10 \quad \text { (taking log both thesides) } \\&\therefore 0.057 t=\ln (10)=2.303 \\&\therefore t=40.39 \mathrm{sec}\end{aligned}

c)

In conclusion, the direction of the induced current will be Counterclockwise.

Read more about current

brainly.com/question/13076734

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4 0
2 years ago
When you drive through deep water, you should dry out your brakes by _____________. pumping your brakes and driving in low gear
Mandarinka [93]

Answer:

By Applying pressure to the brakes

Explanation:

Driving cars through deep water that is more than 10cm can make the cars to float. Most modern cars are usually water- tight so they can start to float through water that is about 30cm deep, fast moving water is very powerful so one needs to be very careful when driving.

If the brakes are wet test them by pressing or tapping on them gently.

You can as well dry brakes by driving in low gear and applying pressure to the brakes.

7 0
3 years ago
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