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dezoksy [38]
2 years ago
10

A student in the front of a school bus tosses a ball to another student in the back of the bus while the bus is moving forward a

t constant velocity. The speed of the ball as seen by a stationary observer in the street:_________
a. is less than that observed inside the bus.
b. is the same as that observed inside the bus
c. may be either greater or smaller than that observed inside the bus.
d. may be either greater, smaller or equal than that observed inside the bus.
e. is greator than that observed inside the bus.
Physics
1 answer:
IRINA_888 [86]2 years ago
8 0

Answer:

d. may be either greater, smaller, or equal to that observed inside the bus.

Explanation:

  • The bus is moving at a constant speed. The ball tossed and received by the ball is inside the bus at a speed equal to the speed of the ball. Therefore the speed of the bus becomes zero with respect to the observer inside the bus.
  • Now the observer inside the bus noticed the ball from the inside of the bus, so he threw the ball back and forth from the ball with the speed v relative to the observer.
  • Now the observer outside the bus could see the bus moving at speed relative to its reference point and also throwing the ball from front to back. The speed of the ball to the observer outside the bus The speed of the bus to the observer outside the bus is minus the speed of the ball to the observer inside the bus.
  • Therefore, the ball speed = (u-v) relative to the observer outside the bus.
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To crossing over the flooded canal.
zubka84 [21]

Answer:

F. jumping

Explanation:

you can't throw/toss yourself, you cant roll over water, catching?, you cant run over water, jumps are bigger than hops

4 0
2 years ago
8.4-1 Consider a magnetic field probe consisting of a flat circular loop of wire with radius 10 cm. The probe’s terminals corres
Vlad1618 [11]

Answer:

B_o = 1.013μT

Explanation:

To find B_o you take into account the formula for the emf:

\epsilon=-\frac{d\Phi_b}{dt}=-\frac{dBAcos\theta}{dt}=-Acos\theta\frac{dB}{dt}

where you used that A (area of the loop) is constant, an also the angle between the direction of B and the normal to A.

By applying the derivative you obtain:

\epsilon=-Acos\theta (2\pi f) B_ocos(2\pi f t+ \alpha)

when the emf is maximum the angle between B and the normal to A is zero, that is, cosθ = 1 or -1. Furthermore the cos function is 1 or -1. Hence:

\epsilon=2\pi fAB_o=2\pi (100*10^3Hz)(\pi (0.1m)^2)B_o=19739.20Hzm^2B_o\\\\B_o=\frac{20*10^{-3}V}{19739.20Hzm^2}=1.013*10^{-6}T=1.013\mu T

hence, B_o = 1.013μT

6 0
2 years ago
Two speedboats are traveling at the same speed relative to the water in opposite directions in a moving river. An observer on th
DENIUS [597]

Answer:

a) vboat = 5.95 m/s  b) vriver= 1.05 m/s

Explanation:

a) As observed from the shore, the speed of the boats can be expressed as the vector sum, of the boat speed relative to the water and the river speed relative to the shore, as follows:

vb₁s = vb₁w + vrs

In one case, the boat is moving in the same direction as the water:

vb₁s = vb₁w + vrs = 7.0 m/s (1)

For the other boat, it is clear that is moving in an opposite direction:

vb₂s = vb₂w - vrs = 4.9 m/s (2)

As  we know that vb₁w = vb₂w, adding both sides, we can remove the river speed from the equation, as follows:

vb₁w = vb₂w =  \frac{7.0 m/s + 4.9 m/s}{2} =5.95 m/s

b) Replacing this value in (1) and solving for vriver, we have:

vriver = 7.0 m/s - 5.95 m/s = 1.05 m/s

(we could have arrived to the same result subtracting both sides in (1), and (2))

3 0
2 years ago
A 5.00μF parallel-plate capacitor is connected to a 12.0 V battery. After the capacitor is fully charged, the battery is disconn
EastWind [94]

(a) 12.0 V

In this problem, the capacitor is connected to the 12.0 V, until it is fully charged. Considering the capacity of the capacitor, C=5.00 \mu F, the charged stored on the capacitor at the end of the process is

Q=CV=(5.00 \mu F)(12.0 V)=60 \mu C

When the battery is disconnected, the charge on the capacitor remains unchanged. But the capacitance, C, also remains unchanged, since it only depends on the properties of the capacitor (area and distance between the plates), which do not change. Therefore, given the relationship

V=\frac{Q}{C}

and since neither Q nor C change, the voltage V remains the same, 12.0 V.

(b) (i) 24.0 V

In this case, the plate separation is doubled. Let's remind the formula for the capacitance of a parallel-plate capacitor:

C=\frac{\epsilon_0 \epsilon_r A}{d}

where:

\epsilon_0 is the permittivity of free space

\epsilon_r is the relative permittivity of the material inside the capacitor

A is the area of the plates

d is the separation between the plates

As we said, in this case the plate separation is doubled: d'=2d. This means that the capacitance is halved: C'=\frac{C}{2}. The new voltage across the plate is given by

V'=\frac{Q}{C'}

and since Q (the charge) does not change (the capacitor is now isolated, so the charge cannot flow anywhere), the new voltage is

V'=\frac{Q}{C'}=\frac{Q}{C/2}=2 \frac{Q}{C}=2V

So, the new voltage is

V'=2 (12.0 V)=24.0 V

(c) (ii) 3.0 V

The area of each plate of the capacitor is given by:

A=\pi r^2

where r is the radius of the plate. In this case, the radius is doubled: r'=2r. Therefore, the new area will be

A'=\pi (2r)^2 = 4 \pi r^2 = 4A

While the separation between the plate was unchanged (d); so, the new capacitance will be

C'=\frac{\epsilon_0 \epsilon_r A'}{d}=4\frac{\epsilon_0 \epsilon_r A}{d}=4C

So, the capacitance has increased by a factor 4; therefore, the new voltage is

V'=\frac{Q}{C'}=\frac{Q}{4C}=\frac{1}{4} \frac{Q}{C}=\frac{V}{4}

which means

V'=\frac{12.0 V}{4}=3.0 V

3 0
2 years ago
It's possible to boil water by adding hot rocks to it, a technique that has been used in many societies over time. If you heat a
Flauer [41]

Answer:

6

Explanation:

m = Mass

c = Specific heat

\Delta T = Change in temperature

s denotes stone

Heat is given by

Q=mc\Delta T\\\Rightarrow Q=5\times 4180\times (100-10)\\\Rightarrow Q=1881000\ J

Heat for the stone

Q=m_sc_s\Delta T\\\Rightarrow 1881000=m_s800\times (500-100)\\\Rightarrow m_s=\dfrac{1881000}{800\times (500-100)}\\\Rightarrow m_s=5.878125\ kg

Number of stones is given by

n=\dfrac{5.878125}{1}\\\Rightarrow n\approx 6

The number of stones is 6

5 0
2 years ago
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