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2 years ago

6

17. How long will it take for an object accelerating at a constant rate of 5 m/s to change its velocity from 0 to 6 m/s? (A) 0.6

s (B) 1.2 s (C) 2.4 s (D) 3.6 s (E) 4.8 s

1 answer:

VikaD [51]2 years ago

8 0

Answer:

(B)

Explanation:

Time = change of velocity ÷ acceleration

= (6-0) ÷ 5

= 1.2

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Twin A makes a one way trip at 0.6c to a star 12 light years away while twin B stays on Earth. Each twin sends the other a signa

hram777 [196]

Answer:

8 signals received by twin A during the trip.

Explanation:

Given that,

Distance = 12 light year

Speed = 0.6 c

Time = 1 year

We need to calculate the time by A

Using formula of time

$T=t\sqrt{\dfrac{1+\dfrac{v}{c}}{1-\dfrac{v}{c}}}$

Put the value into the formula

$T=1\sqrt{\dfrac{1+0.6}{1-0.6}}$

$T=2\ years$

Similarly,

The expression for distance cover by A

$D=d\sqrt{1-\dfrac{v^2}{c^2}}$

$D=12\sqrt{1-(0.6)^2}$

$D=9.6\ ly$

We need to calculate the time

Using formula of time

$t=\dfrac{D}{v}$

$t=\dfrac{9.6}{0.6}$

$t=16\ years$

We need to calculate the signals received by twin A

Using formula for number of signals

$n=\dfrac{t}{T}$

Put the value into the formula

$n=\dfrac{16}{2}$

$n=8\ signals$

Hence, 8 signals received by twin A during the trip.

7 0

3 years ago

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 V. It is used to charge two storage batt

Natali [406]

Complete Question

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 $\Omega$ . It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300 $\Omega$ . If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?

Answer:

a

The additional resistance is $R_z = 4.4 \Omega$

b

The rate at which internal energy increase at the supply is $Z_1 = 32 W$

c

The rate at which internal energy increase in the battery is $Z_1 = 32 W$

d

The rate at which internal energy increase in the added series resistance is $Z_3 = 70.4 W$

e

the increase rate of the chemically energy in the battery is $C = 48 W$

Explanation:

From the question we are told that

The open circuit voltage is $V = 40.0V$

The internal resistance is $R = 2 \Omega$

The emf of each battery is $e = 6.00 V$

The internal resistance of the battery is $r = 0.300V$

The charging current is $I = 4.00 \ A$

Let assume the the additional resistance to to added to the circuit is $R_z$

So this implies that

The total resistance in the circuit is

$R_T = R + 2r +R_z$

Substituting values

$R_T = 2.6 +R_z$

And the difference in potential in the circuit is

$E = V -2e$

=> $E = 40 - (2 * 6)$

$E = 28 V$

Now according to ohm's law

$I = \frac{E}{R_T}$

Substituting values

$4 = \frac{28}{R_z + 2.6}$

Making $R_z$ the subject of the formula

So $R_z = \frac{28 - 10.4}{4}$

$R_z = 4.4 \Omega$

The increase rate of internal energy at the supply is mathematically represented as

$Z_1 = I^2 R$

Substituting values

$Z_1 = 4^2 * 2$

$Z_1 = 32 W$

The increase rate of internal energy at the batteries is mathematically represented as

$Z_2 = I^2 r$

Substituting values

$Z_2 = 4^2 * 2 * 0.3$

$Z_2 = 9.6 \ W$

The increase rate of internal energy at the added series resistance is mathematically represented as

$Z_3 = I^2 R_z$

Substituting values

$Z_3 = 4^2 * 4.4$

$Z_3 = 70.4 W$

Generally the increase rate of the chemically energy in the battery is mathematically represented as

$C = 2 * e * I$

Substituting values

$C = 2 * 6 * 4$

$C = 48 W$

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3 years ago

Assume that the position vector of A is r=i+j+k . Determine the moment about the origin O if the force F=(1)i+(0)j+(5)k . The mo

ddd [48]

Answer:

M₀ = 5i - 4j - k

Explanation:

Using the cross product method, the moment vector(M₀) of a force (F) is about a given point is equal to cross product of the vector A from the point (r) to anywhere on the line of action of the force itself. i.e

M₀ = r x F

From the question,

r = i + j + k

F = 1i + 0j + 5k

Therefore,

M₀ = (i + j + k) x (1i + 0j + 5k)

M₀ = $\left[\begin{array}{ccc}i&j&k\\1&1&1\\1&0&5\end{array}\right]$

M₀ = i(5 - 0) -j(5 - 1) + k(0 - 1)

M₀ = i(5) - j(4) + k(-1)

M₀ = 5i - 4j - k

Therefore, the moment about the origin O of the force F is

M₀ = 5i - 4j - k

3 0

3 years ago

If a sled has a mass of 4kg what is the force of gravity on the sled?

ale4655 [162]

Answer:

Explanation:

Gravity pulls everything down at the same rate of 9.8 m/s/s. If you're looking for the normal force, which is the same as the weight of the object, we'll find that, just in case.

w = mg which says that the normal force/weight of an object is equal to its mass times the pull of gravity:

w = 4.0(9.8) so

w = 39N

7 0

3 years ago

Which statements describe magnetic poles? Check all that apply.

Alex_Xolod [135]

Explanation:

Magnet: It has two poles: South pole and North pole.

Magnetic field lines are stronger near the poles of the magnet.

Same poles repel each other. There is a magnetic force of repulsion between the same poles. North- North poles repel each other.

Unlike poles attract each other. There is magnetic force of attraction between the opposite poles. South- North poles attract each other.

Mono poles cannot exist.

From the given statements, the magnetic poles are described by:

A north pole must exist with a south pole.

Two south poles placed near each other will repel each other.

A north pole and a south pole placed near each other will attract each other.

5 0

3 years ago

Read 2 more answers