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Tanzania [10]
3 years ago
11

PLZZZ HELP

Physics
1 answer:
Doss [256]3 years ago
4 0

Answer/Explanation:

The weight of an object is defined as the force that is exerted due to the gravitational force.

Mathematically, it can be written as :

W = m g

Where

m is the mass of the object

g is the acceleration due to gravity

Also,  

We know that the value of g varies with respect to the location. At the equator, the value of g is less as compared to the poles.

The feature of an object that affects its weight are :

Mass of the object

Location of the object

How much force Earth exerts on the object

You might be interested in
Write the mathemetical relation between work force and displacement​
il63 [147K]

Answer:

Work is measured as the product of force and the displacement in the direction of the force. Work = force × displacement in the direction of the force.

8 0
2 years ago
An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.40 kW. A receiving antenna 6
lara [203]

To solve the problem we will apply the concepts related to the Intensity as a function of the power and the area, as well as the electric field as a function of the current, the speed of light and the permeability in free space, as shown below.

The intensity of the wave at the receiver is

I = \frac{P_{avg}}{A}

I = \frac{P_{avg}}{4\pi r^2}

I = \frac{3.4*10^3}{4\pi(4*1609.34)^2} \rightarrow 1mile = 1609.3m

I = 6.529*10^{-6}W/m^2

The amplitude of electric field at the receiver is

I = \frac{E_{max}^2}{2\mu_0 c}

E_{max}= \sqrt{2I\mu_0 c}

The amplitude of induced emf by this signal between the ends of the receiving antenna is

\epsilon_{max} = E_{max} d

\epsilon_{max} = \sqrt{2I \mu_0 cd}

Here,

I = Current

\mu_0 = Permeability at free space

c = Light speed

d = Distance

Replacing,

\epsilon_{max} = \sqrt{2(6.529*10^{-6})(4\pi*10^{-7})(3*10^{8})(60.0*10^{-2})}

\epsilon_{max} = 0.05434V

Thus, the amplitude of induced emf by this signal between the ends of the receiving antenna is 0.0543V

6 0
4 years ago
The density of a certain type of plastic is 0.75 g/cm3 . If a sheet of this plastic is 10.0 m long, 1.0 m wide, and 1 cm thick,
Oksanka [162]

Answer:

7.5 g

Explanation:

8 0
3 years ago
Determine the approximate force (N) used to pull a sled up a 400 m hill using 1900 J of work.
Sergeu [11.5K]
The work done to pull the sled up to the hill is given by
W=Fd
where
F is the intensity of the force
d is the distance where the force is applied.

In our problem, the work done is W=1900 J and the distance through which the force is applied is d=400 m, so we can calculate the average force by re-arranging the previous equation and by using these data:
F= \frac{W}{d}= \frac{1900 J}{400 m} = 4.75 N \sim 5 N
4 0
4 years ago
A particle (m = 4.3 × 10^-28 kg) starting from rest, experiences an acceleration of 2.4 × 10^7 m/s^2 for 5.0 s. What is its de B
Novay_Z [31]

Answer:

Wavelength, \lambda=1.28\times 10^{-14}\ m

Explanation:

Given that,

Mass of the particle, m=4.3\times 10^{-28}\ kg

Acceleration of the particle, a=2.4\times 10^7\ m/s^2

Time, t = 5 s

It starts from rest, u = 0

The De Broglie wavelength is given by :

\lambda=\dfrac{h}{mv}

v = a × t

\lambda=\dfrac{h}{mat}

\lambda=\dfrac{6.67\times 10^{-34}}{4.3\times 10^{-28}\times 2.4\times 10^7\times 5}

\lambda=1.28\times 10^{-14}\ m

Hence, this is the required solution.

4 0
3 years ago
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