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aleksley [76]
2 years ago
11

A dipole of moment 0.5 e·nm is placed in a uniform electric field with a magnitude of 8 times 104 N/C. What is the magnitude of

the torque on the dipole when (a) the dipole is parallel to the electric field,
Physics
1 answer:
Over [174]2 years ago
8 0

Answer:

The net torque is zero

Explanation:

Let's assume that the dipole is compose of two equal but oposite charges e, and it cam be represented by a rod with one end having a charge e and the other end with a charge of -e. Notice that the dipole is parallel to the electric field thus the force felt by both of the charges will be parallel to the electric field. This means that there will be no components of the forces that are perpendicular to the rod which is a requirement for it to have a torque.

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By means of a rope whose mass is negligible, two blocks are suspended over a pulley, as the drawing shows. The pulley can be tre
tankabanditka [31]

Answer:

Mass of the pull is 77 kg

Explanation:

Here we have for

Since  the rope moves along with pulley, we have

For the first block we have

T₁ - m₁g = -m₁a = -m₁g/4

T₁ =  3/4(m₁g) = 323.4 N

Similarly, as the acceleration of the second block is the same as the first block but in opposite direction, we have

T₂ - m₂g = m₂a = m₂g/4

T₂ =  5/4(m₂g) = 134.75 N

T₂r - T₁r = I·∝ = 0.5·M·r²(-α/r)

∴ M = -\frac{2}{a} (T_2-T_1)

M = -\frac{2}{2.45} (134.75-323.4) = 77 \, kg

Mass of the pull = 77 kg.

5 0
2 years ago
To avoid an accident, a driver steps on the brakes to stop a 1000-kg car traveling at 65km/h. if the braking distance is 35 m, h
IgorLugansk [536]
First, find the needed acceleration needed for the car to stop from its initial velocity given the distance. This is calculated through the equation,
                                             2ad = Vf² - Vi²
where a and d are acceleration and distance, respectively. Vf and Vi are final and initial velocities, respectively. Substituting the known values,
           2(a)(35 m) = (0 m/s)² - ((65 km/h) x (1000 m/ 1 km) x (1 hr / 3600 s))²
The value of acceleration is -4.66 m/s².
The force needed to stop the car is the product of the mass and the acceleration. The operations gives us an answer of -4,660 N. We take the positive value, 4,660 N. 
6 0
2 years ago
Ferdinand the frog is hopping from lily pad to lily pad in search of a good fly
loris [4]

Answer: 36.86\°

Explanation:

According to the described situation we have the following data:

Horizontal distance between lily pads: d=2.4 m

Ferdinand's initial velocity: V_{o}=5 m/s

Time it takes a jump: t=0.6 s

We need to find the angle \theta at which Ferdinand jumps.

In order to do this, we first have to find the <u>horizontal component (or x-component)</u> of this initial velocity. Since we are dealing with parabolic movement, where velocity has x-component and y-component, and in this case we will choose the x-component to find the angle:

V_{x}=\frac{d}{t} (1)

V_{x}=\frac{2.4 m}{0.6 s} (2)

V_{x}=4 m/s (3)

On the other hand, the x-component of the velocity is expressed as:

V_{x}=V_{o}cos\theta (4)

Substituting (3) in (4):

4 m/s=5 m/s cos\theta (5)

Clearing \theta:

\theta=cos^{-1} (\frac{4 m/s}{5 m/s})

\theta=36.86\° This is the angle at which Ferdinand the frog jumps between lily pads

4 0
2 years ago
A raft is made of a plastic block with a density of 650 kg/m 3 , and its dimensions are 2.00 m à 3.00 m à 5.00 m. 1. what is the
cupoosta [38]
1) The volume of the raft is the product between the lenghts of its three dimensions:
V = (2.00 m)(3.00m)(5.00m)=30 m^3

2) The mass of the raft is the product between its density, d, and its volume, V:
m=dV=(650 kg/m^3)(30 m^3)=19500 kg

3) The weight of the raft is the product between its mass m and the gravitational acceleration, g=9.81 m/s^2:
W=mg=(19500 kg)(9.81 m/s^2)=1.91 \cdot 10^5 N

4) The apparent weight is equal to the difference between the weight of the raft and the buoyancy (the weight of the displaced fluid):
W_a = W- \rho_W V_{disp} g
where \rho _W = 1000 kg/m^3 is the water density and V_{disp} is the volume of displaced fluid.
The density of the raft (650 kg/m^3) is smaller than the water density (1000 kg/m^3), this means that initially the buoyancy (which has upward direction) is larger than the weight (downward direction) and so the raft is pushed upward, until it reaches a condition of equilibrium and it floats. At equilibrium, the weight and the buoyancy are equal and opposite in sign:
W=B=\rho _W V_{disp} g
and therefore, the apparent weight will be zero:
W_a = W-B=W-W=0

5) The buoyant force B is the weight of the displaced fluid, as said in step 4):
B=\rho_W V_{disp} g
When the raft is completely immersed in the water, the volume of fluid displaced V_{disp} is equal to the volume of the raft, V_{disp}=V. Therefore the buoyancy in this situation is
B= \rho_W V g = (1000 kg/m^3)(30 m^3)(9.81 m/s^2)=2.94 \cdot 10^5 N
However, as we said in point 4), the raft is pushed upward until it reaches equilibrium and it floats. At equilibrium, the buoyancy will be equal to the weight of the raft (because the raft is in equilibrium), so:
B=W=1.91 \cdot 10^5 N

6) At equilibrium, the mass of the displaced water is equal to the mass of the object. In fact, at equilibrium we have W=B, and this can be rewritten as
mg = m_{disp} g
where m_{disp}= \rho_W V_{disp} is the mass of the displaced water. From the previous equation, we obtain that m_{disp}=m=19500 kg.

7) Since we know that the mass of displaced water is equal to the mass of the raft, using the relationship m=dV we can rewrite m=m_{disp} as:
d V =d_W V_{disp}
and so
V_{disp}= \frac{d V}{d_W}= \frac{(650 kg/m^3)(30m^3)}{1000kg/m^3}= 19.5 m^3

8) The volume of water displaced is (point 7) 19.5 m^3. This volume is now "filled" with part of the volume of the raft, therefore 19.5 m^3 is also the volume of the raft below the water level. We can calculate the fraction of raft's volume below water level, with respect to the total volume of the raft, 30 m^3:
\frac{19.5 m^3}{30 m^3}\cdot 100= 65 \%
Viceversa, the volume of raft above the water level is 30 m^3-19.5 m^3 = 10.5 m^3. Therefore, the fraction of volume of the raft above water level is
\frac{10.5 m^3}{30 m^3}\cdot 100 = 35 \%

9) Let's repeat steps 5-8 replacing \rho _W, the water density, with \rho_E=806 kg/m^3, the ethanol density.

9-5) The buoyant force is given by:
B=\rho _E V_{disp} g = (806 kg/m^3)(30 m^3)(9.81 m/s^2)=2.37 \cdot 10^5 N
when the raft is completely submerged. Then it goes upward until it reaches equilibrium and it floats: in this condition, B=W, so the buoyancy is equal to the weight of the raft.

9-6) Similarly as in point 6), the mass of the displaced ethanol is equal to the mass of the raft: 
m_E = m = 19500 kg

9-7) Using the relationship d= \frac{m}{V}, we can find the volume of displaced ethanol:
V_E =  \frac{m}{d_E} = \frac{19500 kg}{806 kg/m^3}=24.2 m^3

9-8) The volume of raft below the ethanol level is equal to the volume of ethanol displaced: 24.2 m^3. Therefore, the fraction of raft's volume below the ethanol level is 
\frac{24.2 m^3}{30 m^3}\cdot 100 = 81 \%
Consequently, the raft's volume above the ethanol level is 
30 m^3 - 24.2 m^3 = 5.8 m^3
and the fraction of volume above the ethanol level is
\frac{5.8 m^3}{30 m^3}\cdot 100 = 19 \%
8 0
2 years ago
An amusement park ride spins you around in a circle of radius 2.5 m with a speed of 8.5 m/s. If your mass is 75 kg, what is the
zhannawk [14.2K]
B is the answer to this truly did the math
3 0
2 years ago
Read 2 more answers
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