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Lady bird [3.3K]

4 years ago

10

run at the same speed and in the same direction, and they both run for the same amount of time, what can you say about the dista

nce they travel?

1 answer:

LenaWriter [7]4 years ago

4 0

Answer:

the are equivalent

Explanation:

i just learned about that

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(b) A car of mass 3000 kg travels at a constant velocity of 5.0 m/s.

Tamiku [17]

Answer:

16.7 s

Explanation:

T= <u>Vf - Vo</u> a= <u>F</u>

a m

4,500 / 3000 = 1.5 (a)

30 - 5 / 1.5(a) = 16.7 s

4 0

3 years ago

A star with a large luminosity would have a relatively _____ absolute magnitude.

Mademuasel [1]

A star with large luminosity would have a relatively low absolute magnitude. Absolute magnitude is a number that tells how bright a star is from the Earth. However, this scale is backwards and logarithmic, so having a large absolute magnitude value means that the star is faint.

3 0

3 years ago

In order to change a 2.15 kg handcart's velocity from 1.2 m/s

photoshop1234 [79]

Answer: E 1.29 N•s

Explanation: math

7 0

2 years ago

**URGENT** Roberto plans to use two transformers to reduce a voltage of 120 V to 4 V. He uses a transformer that has 300 coils i

skelet666 [1.2K]

As we know that in transformers we have

$\frac{V_s}{V_p} = \frac{N_s}{N_p}$

here we know that

$V_s = 4 Volts$

$V_p = 120 Volts$

$N_s = 50 coils$

$N_p = 300 coils$

now from above equation we will have

$\frac{V}{120} = \frac{50}{300}$

$V = 20 Volts$

now we have to reduce this voltage to final voltage of V = 4 V

so again we will have

$\frac{V_s}{V_p} = \frac{N_s}{N_p}$

$\frac{4}{20} = \frac{N_s}{N_p}$

$\frac{N_s}{N_p} = \frac{1}{5}$

so we need to take such a winding whose ratio is 1:5

So it is satisfied in X

$N_p = 60$

$N_s = 12$

so answer will be

<u>B)- X</u>

3 0

3 years ago

Read 2 more answers

you spray your sister with water from a garden hose. the water is supplied to the hose at a rate of 0.649x10-3 m^3/s and the dia

Ivanshal [37]

Answer:

27.82 m/s

Explanation:

The radius of the hose is half of its diameter

$r = d/2 = 5.45\times10^{-3}/2 = 0.002725 m$

So its area must be

$A = \pi r^2 = \pi 0.002725^2 = 2.33\times10^{-5} m^2$

The speed of water coming out of the hose is its flow rate divided by the cross-section area of the hose

$v = \dot{V}/A =0.000649 / 2.33\times10^{-5} = 27.82 m/s$

7 0

3 years ago