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3 years ago

If a circling object is released, centrifugal force will make it travel away from the center of its original path.

1 answer:

5 0

The answer is True. The object is thrown away from the center of its original path due its inertia. Centrifugal force acts in the opposite direction as centripetal force. Centripetal force applies towards the center of the curvature in a spinning object. Centrifugal force is considered an apparent force while centripetal force is an actual force.

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Special relativity can be used to study an object in which frame of reference?

Tresset [83]

Special relativity led the path for general relativity; special relativity is in a sense a special application of the rules of general relativity. While general relativity is in position to tackle all of these problems, special relativity can tackle only problems in inertial frames. Inertial frame means that the frame of reference is inot accelerating. So, we disqualify answers A and D. However, remember that moving in a circle means that there is an acceleration, the centrifugal one, even if the speed does not change. Hence C is also incorrect.
The correct answer is B, since if there is no change in velocity, the frame does not accelerate and it is inertial.

3 0

3 years ago

Read 2 more answers

A radio station's channel, such as 100.7 fm or 92.3 fm, is actually its frequency in megahertz (mhz), where 1mhz=106 hz and 1hz=

AURORKA [14]

The frequency of the radio station is
$f=88.7 fm= 88.7 MHz = 88.7 \cdot 10^6 Hz$

For radio waves (which are electromagnetic waves), the relationship between frequency f and wavelength $\lambda$ is
$\lambda= \frac{c}{f}$
where c is the speed of light. Substituting the frequency of the radio station, we find the wavelength:
$\lambda= \frac{3 \cdot 10^8 m/s}{88.7 \cdot 10^6 Hz}=3.38 m$

6 0

4 years ago

Read 2 more answers

System A has masses m and m separated by a distance r; system B has masses m and 2m separated by a distance 2r; system C has mas

Anna [14]

Answer:

System D --> System C --> System A --> System B

Explanation:

The gravitational force between two masses m1, m2 separated by a distance r is given by:

$F=G \frac{m_1 m_2}{r^2}$

where G is the gravitational constant. Let's apply this formula to each case now to calculate the relative force for each system:

System A has masses m and m separated by a distance r:

$F=G\frac{m \cdot m}{r^2}=G \frac{m^2}{r^2}$

system B has masses m and 2m separated by a distance 2r:

$F=G\frac{m \cdot 2m}{(2r)^2}=G \frac{2m^2}{4r^2}=\frac{1}{2} G \frac{m^2}{r^2}$

system C has masses 2m and 3m separated by a distance 2r:

$F=G\frac{2m \cdot 3m}{(2r)^2}=G \frac{6m^2}{4r^2}=\frac{3}{2} G \frac{m^2}{r^2}$

system D has masses 4m and 5m separated by a distance 3r:

$F=G\frac{4m \cdot 5m}{(3r)^2}=G \frac{20m^2}{9r^2}=\frac{20}{9} G \frac{m^2}{r^2}$

Now, by looking at the 4 different forces, we can rank them from the greatest to the smallest force, and we find:

System D --> System C --> System A --> System B

5 0

3 years ago

B. Study the given figure and calculate the area of the leaf

yanalaym [24]

Answer:

19 cm sq

Explanation:

you have to count one full and half area occupied by leaf in each box, then you have to multiply with 1 cm sq.

4 0

3 years ago

A laboratory technician drops a 72.0 g sample of unknown solid material, at a temperature of 80.0°C, into a calorimeter. The cal

Natalija [7]

Answer : The specific heat of unknown sample is, $8748.78J/kg^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-[q_2+q_3]$

$m_1\times c_1\times (T_f-T_1)=-[m_2\times c_2\times (T_f-T_2)+m_3\times c_3\times (T_f-T_2)]$

where,

$c_1$ = specific heat of unknown sample = ?

$c_2$ = specific heat of water = $4186J/kg^oC$

$c_3$ = specific heat of copper = $390J/kg^oC$

$m_1$ = mass of unknown sample = 72.0 g = 0.072 kg

$m_2$ = mass of water = 203 g = 0.203 kg

$m_2$ = mass of copper = 187 g = 0.187 kg

$T_f$ = final temperature of calorimeter = $39.4^oC$

$T_1$ = initial temperature of unknown sample = $80.0^oC$

$T_2$ = initial temperature of water and copper = $11.0^oC$

Now put all the given values in the above formula, we get

$0.072kg\times c_1\times (39.4-80.0)^oC=-[(0.203kg\times 4186J/kg^oC\times (39.4-11.0)^oC)+(0.187kg\times 390J/kg^oC\times (39.4-11.0)^oC)]$

$c_1=8748.78J/kg^oC$

Therefore, the specific heat of unknown sample is, $8748.78J/kg^oC$

7 0

3 years ago