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NARA [144]
2 years ago
15

Falls often cause injuries, so one of the significant aspects of falls is the displacement and motion of a body. It is defined a

s how far a body moves vertically during the fall. Accordingly, there are three important factors that should be known to compute the kinetic energy (KE) of a falling body. These factors are:
Physics
1 answer:
Dima020 [189]2 years ago
7 0

Answer:

We need, mass, gravity and height.

Explanation:

When a body falls freely from a height, its initial velocity is zero, but due to the height it has some potential energy at the top and the kinetic energy is zero.

As it falls, the potential energy is gradually converted in to the kinetic energy so that the total energy of the falling body is conserved.

At the time as the body strikes the ground, the entire potential energy is converted into the kinetic energy.

Potential energy is given by

U = m g h

where, m is the mass, g is the gravity and h is the height,

So, to get the kinetic energy we require mass, gravity and height of the body.

You might be interested in
Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m
alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = 0.407\times 10^{-3}\ m

Energy stored in the capacitor (U) = 7.87\ nJ=7.87\times 10^{-9}\ J

Assuming the medium to be air.

So, permittivity of space (ε) = 8.854\times 10^{-12}\ F/m

Area of the square plates is given as:

A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2

Capacitance of the capacitor is given as:

C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F

Now, we know that, the energy stored in a parallel plate capacitor is given as:

U=\frac{CE^2d^2}{2}

Rewriting in terms of 'E', we get:

E=\sqrt{\frac{2U}{Cd^2}}

Now, plug in the given values and solve for 'E'. This gives,

E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0
3 years ago
Suppose a cart with no fans has a starting velocity of 2 m/s. What will be the velocity of the cart when it reaches the wall?
Roman55 [17]

Answer:

less than stating velocity due to friction and air resistance.

Explanation:

3 0
2 years ago
In the following figure, if AB ǁ CD, then find the measure of PCD and CPD.​
mina [271]

Answer:

CPD = 80

PCD = 44

Explanation:

Given

AB || CD

BAD = 56

CPA = 100

See attachment

Required

Determine PCD and CPD

First, we need to calculate CPD

Since DPA is a straight line and CPA = 100;

We have that:

CPA + CPD = 180 --- angle on a straight theorem

Substitute 100 for CPA

100 + CPD = 180

Subtract 100 from both sides

100-100 + CPD = 180-100

CPD = 80

Next, we calculate PCD

We have that:

DAB= ADC = 56  --alternate angle

In triangle PCD

PCD + CPD + PDC = 180 --- angles in a triangle

Where

PDC = ADC = 56

So, we have:

PCD +80 + 56 = 180

PCD +136 = 180

Subtract 136 from both sides

PCD = 180 - 136

PCD = 44

6 0
3 years ago
A long, straight, horizontal wire carries a left-to-right current of 40 A. If the wire is placed in a uniform magnetic field of
sladkih [1.3K]

Answer:

The magnitude of the resultant of the magnetic field is 4.11\times10^{-5}\ T

Explanation:

Given that,

Current = 40 A

Magnetic field B=3.7\times10^{-5}\ T

Distance = 22 cm

We need to calculate the magnetic field

Using formula of magnetic field

B'=\dfrac{\mu_{0}I}{2\pi r}

Where, r = distance

I = current

Put the value into the formula

B'=\dfrac{4\pi\times10^{-7}\times20}{2\pi\times0.22}

B'=1.8\times10^{-5}\ T

We need to calculate the magnitude of the resultant of the magnetic field

Using formula of resultant

B''=\sqrt{B^2+B'^2}

Put the value into the formula

B''=\sqrt{(3.7\times10^{-5})^2+(1.8\times10^{-5})^2}

B''=4.11\times10^{-5}\ T

Hence, The magnitude of the resultant of the magnetic field is 4.11\times10^{-5}\ T

6 0
3 years ago
A boy drags a suitcase along the ground with a force of 100 N. If the frictional force opposing the motion of the suitcase is 50
stira [4]
Fortunately, 'force' is a vector.  So if you know the strength and direction
of each force, you can easily addum up and find the 'resultant' (net) force.

When we talk in vectors, one newton forward is the negative of
one newton backward.   Hold that thought, while I slog through
the complete solution of the problem.


            (100 N forward) plus (50 N backward)

        =  (100 N forward) minus (50 N forward)

        =           50 N forward .

That's it.
Is there any part of the solution that's not clear ?

4 0
3 years ago
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