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-Dominant- [34]
3 years ago
6

What was Einstein's theory of relativity?

Physics
1 answer:
Verizon [17]3 years ago
8 0
He determined that the laws of physics are the same for all non-accelerating observers <span />
You might be interested in
The electrons in the beam of a television tube have a kinetic energy of 2.20 10-15 j. initially, the electrons move horizontally
dalvyx [7]
(a) The electrons move horizontally from west to east, while the magnetic field is directed downward, toward the surface. We can determine the direction of the force on the electron by using the right-hand rule:
- index finger: velocity --> due east
- middle finger: magnetic field --> downward
- thumb: force --> due north
However, we have to take into account that the electron has negative charge, therefore we have to take the opposite direction: so, the magnetic force is directed southwards, and the electrons are deflected due south.

b) From the kinetic energy of the electrons, we can find their velocity by using
K= \frac{1}{2}mv^2
where K is the kinetic energy, m the electron mass and v their velocity. Re-arranging the formula, we find
v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 2.20 \cdot 10^{-15} J}{9.1 \cdot 10^{-31} kg} }=6.95 \cdot 10^7 m/s

The Lorentz force due to the magnetic field provides the centripetal force that deflects the electrons:
qvB = m \frac{v^2}{r}
where
q is the electron charge
v is the speed
B is the magnetic field strength
m is the electron mass
r is the radius of the trajectory
By re-arranging the equation, we find the radius r:
r= \frac{mv}{qB}= \frac{(9.1 \cdot 10^{-31} kg)(6.95 \cdot 10^7 m/s)}{(1.6 \cdot 10^{-19} C)(3.00 \cdot 10^{-5} T)}=13.18 m

And finally we can calculate the centripetal acceleration, given by:
a_c =  \frac{v^2}{r}= \frac{(6.95 \cdot 10^7 m/s)^2}{13.18 m}=3.66 \cdot 10^{14} m/s^2
5 0
3 years ago
3. A football is kicked with a speed of 35 m/s at an angle of 40°.
jarptica [38.1K]

a) 22.5 m/s

The initial vertical velocity is given by:

u_y = u sin \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_y = (35)(sin 40)=22.5 m/s

b) 26.8 m/s

The initial horizontal velocity is given by:

u_x = u cos \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_x = (35)(cos 40)=26.8 m/s

c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

v_y = u_y + at

where

v_y is the vertical velocity at time t

u_y = 22.5 m/s

a=g=-9.8 m/s^2 is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, v_y =0; substituting, we find the time t at which this happens:

0=u_y + gt\\t=-\frac{u_y}{g}=-\frac{22.5}{-9.8}=2.30 s

d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

s=u_y t + \frac{1}{2}gt^2

where

s is the vertical displacement at time t

u_y = 22.5 m/s

g=-9.8 m/s^2

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

t=2(2.30 s)=4.60 s

Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

v_x = 26.8 m/s

Therefore, the horizontal distance travelled during the whole motion is

d=v_x t = (26.8)(4.60)=123.3 m

So, the ball lands 123.3 m far from the initial point.

4 0
3 years ago
When water freezes, its volume increases by 9.05%. What force per unit area is water capable of exerting on a container when it
AnnZ [28]

Answer:

The Pressure is 0.20 MPa.

(b) is correct option.

Explanation:

Given that,

Change in volume = 9.05%

{tex]\dfrac{\Delta V}{V_{0}}=0.0905[/tex]

We know that.

The bulk modulus for water

B=0.20\times10^{10}\ N/m^2

We need to calculate the pressure difference

Using formula bulk modulus formula

B=\Delta P\dfrac{V_{0}}{\Delta V}

\Delta P=B\dfrac{\Delta V}{V_{0}}

\Delta P=0.2\times10^{10}\times0.0905

\Delta P=0.2\times10^{6}\ Pa

\Delta P=0.20 MPa

Hence, The Pressure is 0.20 MPa.

6 0
4 years ago
What is the difference between an induced and a permanent magnet?
MissTica
INDUCTION MOTOR:-

Speed:-Less speed range than PMAC motors • Speed range is a function of the drive being used — to 1,000:1 with an encoder, 120:1 under field-oriented control


Reliability:-Waste heat is capable of degrading insulation essential to motor operation • Years of service common with proper operation

Power density:-Induction produced by squirrel cage rotor inherently limits power density

Accuracy:-Flux vector and field-oriented control allows for some of accuracy of servos

Cost:-Relatively modest initial cost; higher operating costs

PERMANENT MAGNET MORTOR:-

speed:-VFD-driven PMAC motors can be used in nearly all induction-motor and some servo applications • Typical servomotor application speed — to 10,000 rpm — is out of PMAC motor range

Reliability:-Lower operating temperatures reduces wear and tear, maintenance • Extends bearing and insulation life • Robust construction for years of trouble-free operation in harsh environments.

power density:-Rare-earth permanent magnets produce more flux (and resultant torque) for their physical size than induction types.

Accuracy:-Without feedback, can be difficult to locate and position to the pinpoint accuracy of servomotors

<span>Cost:-Exhibit higher efficiency, so their energy use is smaller and full return on their initial purchase cost is realized more quickly</span>
8 0
3 years ago
If one of the masses of the Atwood's machine below is 2.9 kg, what should be the other mass so that the displacement of either m
Afina-wow [57]

Answer:

2.59 Kg, 3.25 Kg

Explanation:

Acceleration can be found using equation

s=0.5at^{2} where s is the release distance, a is acceleration and t is time

Making a the subject of the formula

a=\frac {2S}{t^{2}}

Substituting 0.28 for s and time for 1 second

a=\frac {2*0.28m}{(1s)^{2}=0.56 m/s^{2}

Acceleration formula for the Atwood machine is given by

a=\frac {g(m1-m2)}{m1+m2}  where m1 and m2 are first and second masses respectively

Two situations are possible

When m1>m2

Assuming m1 is 3.7kg which is heavier than m2

Substituting a for 0.56 m/s^{2}  and m1 as 2.9Kg and taking acceleration due to gravity as 9.81 m/s^{2}  

0.56 m/s^{2}=\frac {9.81 m/s^{2}*(2.9Kg-m2)}{2.9Kg+m2}  

(0.56 m/s^{2})*(2.9Kg+m2)=9.81(2.9Kg-m2)  

(0.56 m/s^{2}+9.81)*m2=(9.81*2.9)-(2.9*0.56)  

10.37m2=28.449-1.624

10.37m2=26.825

m2=\frac {26.825}{10.37}=2.5867888138  

m2=2.59 Kg

<u>When m1<m2</u>

a=\frac {g(m2-m1)}{m1+m2}  

Then m1=2.9Kg hence

0.56 m/s^{2}=\frac {9.81(m2-2.9Kg)}{2.9Kg+m2}

(0.56 m/s^{2})*(2.9Kg+m2)=9.81 m/s^{2}*(m2-2.9Kg)

(0.56 m/s^{2}-9.81 m/s^{2})*m2=(-2.9 Kg*9.81 m/s^{2})-(2.9 Kg*0.56 m/s^{2})  

-9.25m2=-28.449-1.624=-30.073

9.25m2=30.073

m2=\frac {30.073}{9.25}=3.2511351351

m2=3.25 Kg

8 0
3 years ago
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