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mr_godi [17]
2 years ago
10

At t1, Car A and car B are each located at position xo moving forward at speed v. At t2, car A is located at position 2xo moving

forward at speed 3v, while car B is located at position 2xo but is moving backward at speed v. Is the average velocity of car A between t1 and t2greater than, less than, or equal to the average velocity of car B between t1 and t2?
Physics
1 answer:
wel2 years ago
3 0

Answer:

the average velocity of car A between t1 and t2greater is greater than the average velocity of B berween t1 and t2

Explanation:

Velocity is displacement over time,

Displacement is the distance covered relative to the initial starting position

For A:

at time ti, A moved from Xo to 2Xo, displacement is 2Xo.

at time t2 a moves with speed 3V, hence, his new position will be 3Xo from 2Xo which will be at 5Xo. A's displacement is 5Xo from starting point.

For B:

at time ti, B moved from Xo to 2Xo, displacement is 2Xo.

at time t2 a moves with speed V in the opposite position so he'll be back to his starting point, hence, his new position will be at Xo. A's displacement is 0 from his starting point.

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Two convex thin lenses with focal lengths 10.0 cm and 20.0 cm are aligned on a common axis, running left to right, the 10-cm len
love history [14]

Answer:

(c) +6.67

Explanation:

f1 = 10 cm

f2 = 20 cm

u = Object distance = 15 cm

Distance between lenses = 20 cm

For first lens image distance

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{15}\\\Rightarrow \frac{1}{v}=\frac{1}{30}\\\Rightarrow v=30\ cm

Distance from second lens is 10 cm to the right

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{20}-\frac{1}{-10}\\\Rightarrow \frac{1}{v}=\frac{3}{20}\\\Rightarrow v=6.67\ cm

The final image will appear as +6.67 cm

3 0
2 years ago
Case 1: A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t₁ = 11.9 seconds to get
olga_2 [115]

Answer:

Part a)

\omega = 8.17 rad/s

Part b)

N = 7.74 rev

Part c)

\alpha = 0.69 rad/s^2

Part d)

\alpha = 0.48 rad/s^2

Part e)

t = 9.14 s

Explanation:

Part a)

Angular speed is given as

\omega = 2\pi f

\omega = 2\pi(\frac{78}{60})

\omega = 8.17 rad/s

Part b)

Since turn table is accelerating uniformly

so we will have

\theta = \frac{\omega_f + \omega_i}{2} t

\theta = \frac{8.17 + 0}{2}(11.9)

2N\pi = 48.6

N = 7.74 rev

Part c)

angular acceleration is given as

\alpha = \frac{\omega_f - \omega_i}{t}

\alpha = \frac{8.17 - 0}{11.9}

\alpha = 0.69 rad/s^2

Part d)

When its angular speed changes to 120 rpm

then we will have

\omega_2 = 2\pi (\frac{120}{60})

\omega_2 = 12.56 rad/s

number of turns revolved is 15 times

so we have

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

12.56^2 - 8.17^2 = 2\alpha (2\pi\times 15)

\alpha = 0.48 rad/s^2

Part e)

now for uniform acceleration we have

\omega_f - \omega_i = \alpha t

12.56 - 8.17 = 0.48 t

t = 9.14 s

7 0
2 years ago
Based on the article “Will the real atomic model please stand up?,” why did J.J. Thomson experiment with cathode ray tubes? to s
PIT_PIT [208]

Answer:

B.) to determine that electric beams in cathode ray tubes were actually made of particles

Explanation:

This is the right answer i just took the quiz on edge.

3 0
2 years ago
A particle with a charge of 3.00 elementary charges moves through a potential difference of 4.50 volts. What is the change in el
ollegr [7]
The answer:
the relationship between elementary charge, potential difference and electrical potential energy is given by 
   E= qV
E:  lectrical potential energy
q:   elementary charge
V:   potential difference

but we have  e=abs val(q)=3 
so we have E= qV=3ex4.5V=<span>13.5 eV
</span>
the answer is <span>(4)13.5 eV</span>

8 0
2 years ago
Calculate the change in the kinetic energy (KE) of the bottle when the mass is increased. Use the formula KE = mv2, where m is t
Aliun [14]

kinetic energy is given as

KE = (0.5) m v²

given that : v = speed of the bottle in each case =  4 m/s

when m = 0.125 kg

KE = (0.5) m v² =  (0.5) (0.125) (4)² = 1 J

when m = 0.250 kg

KE = (0.5) m v² =  (0.5) (0.250) (4)² = 2 J

when m = 0.375 kg

KE = (0.5) m v² =  (0.5) (0.375) (4)² = 3 J

when m = 0.0.500 kg

KE = (0.5) m v² =  (0.5) (0.500) (4)² = 4 J

6 0
2 years ago
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