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mamaluj [8]
3 years ago
5

What is a gravitational field and how its strength be measured

Physics
1 answer:
Yakvenalex [24]3 years ago
5 0
A gravitational field is the field generated by a massive body, that extends into the entire space. Every object with mass m experiences a force F when immersed in a gravitational field. The intensity of the force is equal to
F= \frac{GM}{r^2}  m
where G=6.67 \cdot 10^{-11} m^3 Kg^{-1} s^{-2} is the gravitational  constant, M is the mass of the source of the field (e.g. the mass of a planet), and r is the distance between the object and the source of the field. The force is always attractive. 

A possible way to measure the intensity of a gravitational field is by measuring the acceleration a of the object immersed in this field. In fact, for Newton's second law we have:
F=ma
but since 
F= \frac{GM}{r^2} m
we can write
a =  \frac{GM}{r^2}
Therefore, by measuring the acceleration of the object, we also measure the intensity of the field.

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(a) What is the minimum width of a single slit (in multiples of λ ) that will produce a first minimum for a wavelength λ ? (b) W
Kitty [74]

Answer:

The minimum value of width for first minima is λ

The  minimum value of width for 50 minima is 50λ

The  minimum value of width for 1000 minima is 1000λ

Explanation:

Given that,

Wavelength = λ

For D to be small,

We need to calculate the minimum width

Using formula of minimum width

D\sin\theta=n\lambda

D=\dfrac{n\lambda}{\sin\theta}

Where, D = width of slit

\lambda = wavelength

Put the value into the formula

D=\dfrac{n\lambda}{\sin\theta}

Here, \sin\theta should be maximum.

So. maximum value of \sin\theta is 1

Put the value into the formula

D=\dfrac{1\times\lambda}{1}

D=\lambda

(b). If the minimum number  is 50

Then, the width is

D=\dfrac{50\times\lambda}{1}

D=50\lambda

(c). If the minimum number  is 1000

Then, the width is

D=\dfrac{1000\times\lambda}{1}

D=1000\lambda

Hence, The minimum value of width for first minima is λ

The  minimum value of width for 50 minima is 50λ

The  minimum value of width for 1000 minima is 1000λ

4 0
3 years ago
A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.25 kg to a friend standing in front of him
juin [17]

Answer:

The velocity of the student has after throwing the book is 0.0345 m/s.

Explanation:

Given that,

Mass of book =1.25 kg

Combined mass = 112 kg

Velocity of book = 3.61 m/s

Angle = 31°

We need to calculate the magnitude of the velocity of the student has after throwing the book

Using conservation of momentum along horizontal  direction

m_{b}v_{b}\cos\theta= m_{c}v_{c}

v_{s}=\dfrac{m_{b}v_{b}\cos\theta}{m_{c}}

Put the value into the formula

v_{c}=\dfrac{1.25\times3.61\times\cos31}{112}

v_{c}=0.0345\ m/s

Hence, The velocity of the student has after throwing the book is 0.0345 m/s.

3 0
2 years ago
Which units are used to express kinetic energy?
labwork [276]

Answer:

The SI units for energy is Joules.

4 0
3 years ago
Read 2 more answers
Which of the following options is correct and why?
Dimas [21]

Answer:

Option (e) = The charge can be located anywhere since flux does not depend on the position of the charge as long as it is inside the sphere.

Explanation:

So, we are given the following set of infomation in the question given above;

=> "spherical Gaussian surface of radius R centered at the origin."

=> " A charge Q is placed inside the sphere."

So, the question is that if we are to maximize the magnitude of the flux of the electric field through the Gaussian surface, the charge should be located where?

The CORRECT option (e) that is " The charge can be located anywhere since flux does not depend on the position of the charge as long as it is inside the sphere." Is correct because of the reason given below;

REASON: because the charge is "covered" and the position is unknown, the flux will continue to be constant.

Also, the Equation that defines Gauss' law does not specify the position that the charge needs to be located, therefore it can be anywhere.

6 0
3 years ago
An automobile engine can produce 153 N · m of torque. Calculate the angular acceleration (in rad/s^2) produced if 85.2% of this
galina1969 [7]

Answer:

46.2 rad/s2

Explanation:

Angular acceleration works very similar to linear acceleration, it follows this equation:

\gamma = \frac{Mt}{J}

Where:

γ: angular acceleration

Mt: torque

J: moment of inertia of the load from its turning axis

Since we have the torque we just need the moment of inertia. We have to add together the moments of the drive shaft, tires, wheel walls and wheels.

The wheels act like disks. For disks the moment of inertia is:

J = \frac{1}{2} * m * r^2

Jwheel = \frac{1}{2} = 15 * 0.18^2 = 0.243 kg*m^2

The wheel walls act like annular rings, for these the moment of inertia is:

J = \frac{1}{2} * m * (re^2 - ri^2)

Jwall = \frac{1}{2} * 2 * (0.32^2 - 0.18^2) = 0.07 kg * m^2

The tread acts like a hoop, as in mass concentrated into a circunference, for these:

J = m * r^2

Jtread = 10 * 0.33^2 = 1.09 kg*m^2

The axle acts like a rod, which is the same as the disk:

Jaxle = \frac{1}{2} * 14.1 * 0.02^2 = 0.0028 kg*m^2

The drive shaft acts like a rod too:

Jshaft = \frac{1}{2} * 31.7 * 0.032^2 = 0.016 kg*m^2

SO, the total moment of inertia is:

J = 2*Jwheel + 2*Jwall + 2*Jtread + Jaxle + Jshaft

J = 2*0.243 + 2*0.07 + 2*1.09 + 0.0028 + 0.016 = 2.82 kg*m2

Finally the angular acceleration is:

\gamma = \frac{0.852 * 153}{2.82} = 46.2 \frac{rad}{s^2}

4 0
3 years ago
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