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KATRIN_1 [288]
3 years ago
6

A rigid rod of length L rotates about an axis perpendicular to the rod, with one end of the rod fixed to the axis. Which of the

following must always be equal at all points on the rod?
1. Angular position.
2. Angular velocity.
3. Angular acceleration.
4. Tangential acceleration.
A. 1 and 2.
B. 1, 2 and 4.
C. 1 and 3.
D. 1, 2, 3 and 4.
Physics
1 answer:
bonufazy [111]3 years ago
3 0

Answer:

D

Explanation:

The Angular Position: The angular position is also called the angular displacement, and it is the angle through which a point revolves around the centre line that has been rotated in a particular manner about a given axis.

The Angular Velocity: The angular velocity of an object is the measure of the speed, or say, how fast an object rotates at one point with respect to a point.

The Angular Acceleration: This is very simple as it's a derivative of angular velocity. They both share the same relationship velocity and acceleration share. Angular acceleration is the rate of change of angular velocity.

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the ocean floor is, on average, 4267 m below sea level. What is the pressure in the atmosphere at this depth?
hichkok12 [17]
The pressure at the depth h in the ocean is given by (Stevin's law)
p= p_0 + \rho g h
where
p_0 = 1.0 \cdot 10^5 Pa is the atmospheric pressure
and \rho g h is the pressure exerted by the column of water of height h=4267 m, with \rho = 1000 kg/m^3 being the water density and g=9.81 m/s^2.
Substituting, we find
p=1.0 \cdot 10^5 Pa + (1000 kg/m^3)(9.81 m/s^2)(4267 m)=4.20 \cdot 10^7 Pa
We want to convert this into atmospheres: we know that 1 atm corresponds to the atmospheric pressure at sea level, so 1 atm=1.0\cdot 10^5 Pa, therefore we just need to divide by this number:
p= \frac{4.20 \cdot 10^7 Pa}{1.0 \cdot 10^5 Pa/atm} =420 atm
7 0
3 years ago
You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics
olga nikolaevna [1]

Answer:112.82 m/s

Explanation:

Given

range of arrow=68 m

Angle=3^{\circ}

as the arrow travels it acquire a vertical velocity v_y

v_y=u+at

v_y=0+9.81\times t-------1

Range is given by

R=ut

where u=initial velocity

68=u\times t

t=\frac{68}{u}

substitute the value of t in eqn 1

v_y=9.81\times \frac{68}{u}

v_y\times u=9.81\times 68=667.08--------2

and tan(3)=\frac{v_y}{u}

v_y=utan(3)=0.0524u

substitute it in 2

0.0524 u^2=667.08

u^2=12,728.644

u=112.82 m/s

6 0
4 years ago
A temperature of 20°C is equal to ? °F.
Papessa [141]
The answer is 68 F. i hope this helps
6 0
4 years ago
Read 2 more answers
A train leaves the train station at noon and travels at a constant speed of vt = 50 mi/hr on a straight track. 2 hr later, a car
asambeis [7]

Answer:

350 miles

Explanation:

When the car starts 2 hours later, the train would have a head start of

50 * 2 = 100 miles

The speed of the car relative to the train is

70 - 50 = 20 mi/hr

For the car to catch up with the train, it must cover the 100 miles difference at the rate of 20mi/hr. So the time it would need to cover this difference is

100 / 20 = 5 hours

After 5 hours, the car would have traveled a distance of

5 * 70 = 350 miles which is also the distance from the station to where the car catches up

6 0
4 years ago
A short current element dl⃗ = (0.500 mm)j^ carries a current of 5.40 A in the same direction as dl⃗ . Point P is located at r⃗ =
SashulF [63]

Answer:

The magnetic field along x axis is

B_{x}=1.670\times10^{-10}\ T

The magnetic field along y axis is zero.

The magnetic field along z axis is

B_{z}=3.484\times10^{-10}\ T

Explanation:

Given that,

Length of the current element dl=(0.5\times10^{-3})j

Current in y direction = 5.40 A

Point P located at \vec{r}=(-0.730)i+(0.390)k

The distance is

|\vec{r}|=\sqrt{(0.730)^2+(0.390)^2}

|\vec{r}|=0.827\ m

We need to calculate the magnetic field

Using Biot-savart law

B=\dfrac{\mu_{0}}{4\pi}\timesI\times\dfrac{\vec{dl}\times\vec{r}}{|\vec{r}|^3}

Put the value into the formula

B=10^{-7}\times5.40\times\dfrac{(0.5\times10^{-3})\times(-0.730)i+(0.390)k}{(0.827)^3}

We need to calculate the value of \vec{dl}\times\vec{r}

\vec{dl}\times\vec{r}=(0.5\times10^{-3})\times(-0.730)i+(0.390)k

\vec{dl}\times\vec{r}=i(0.350\times0.5\times10^{-3}-0)+k(0+0.730\times0.5\times10^{-3})

\vec{dl}\times\vec{r}=0.000175i+0.000365k

Put the value into the formula of magnetic field

B=10^{-7}\times5.40\times\dfrac{(0.000175i+0.000365k)}{(0.827)^3}

B=1.670\times10^{-10}i+3.484\times10^{-10}k

Hence, The magnetic field along x axis is

B_{x}=1.670\times10^{-10}\ T

The magnetic field along y axis is zero.

The magnetic field along z axis is

B_{z}=3.484\times10^{-10}\ T

7 0
3 years ago
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