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2 years ago

15. What is one way in which scientists study dark

Physics

1 answer:

Kay [80]2 years ago

8 0

Answer:

It's d

Explanation:

Scientists study dark matter by looking at how it affects visible matter.

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A 12.0N force with a fixed orientation does work on a

kvasek [131]

Answer:

(a) $\theta=62.31^{\circ}$

(b) $\theta=117.68^{\circ}$

Explanation:

It is given that,

Force acting on the particle, F = 12 N

Displacement of the particle, $d=(2.00i -4.00j+3.00k)\ m$

Magnitude of displacement, $d=\sqrt{2^2+4^2+3^2}= 5.38\ m$

(a) If the change in the kinetic energy of the particle is +30 J. The work done by the particle is given by :

$W=Fd\ cos\theta$

$\theta$ is the angle between force and the displacement

According to work energy theorem, the charge in kinetic energy of the particle is equal to the work done.

So,

$cos\theta=\dfrac{W}{Fd}$

$cos\theta=\dfrac{+30\ J}{12\times 5.38}$

$\theta=62.31^{\circ}$

(b) If the change in the kinetic energy of the particle is (-30) J. The work done by the particle is given by :

$cos\theta=\dfrac{W}{Fd}$

$cos\theta=\dfrac{-30\ J}{12\times 5.38}$

$\theta=117.68^{\circ}$

Hence, this is the required solution.

8 0

3 years ago

What could be the plot of a story about planets/ astronomy?

Finger [1]

Answer:

Some stars in space can come to life causing humans to be amazed but at thesame time in great/critical danger.

3 0

2 years ago

Read 2 more answers

Mass is a measurement of

lozanna [386]

D is the amount of space object takes up

6 0

3 years ago

14. Which of the following is not an example of work being done?

Natali5045456 [20]

Answer:

B. holding a coffee mug

Explanation:

Something must move a distance for work to be done.

7 0

2 years ago

An automobile battery has an emf of 12.6 V and an internal resistance of 0.0600 . The headlights together present equivalent res

murzikaleks [220]

Answer:

(a) $V=11.86\ V$

(b) $V=9.76\ V$

Explanation:

<u>Electric Circuits</u>

Suppose we have a resistive-only electric circuit. The relation between the current I and the voltage V in a resistance R is given by the Ohm's law:

$V=R.I$

(a) The electromagnetic force of the battery is $\varepsilon =12.6\ V$ and its internal resistance is $R_i=0.06\ \Omega$ . Knowing the equivalent resistance of the headlights is $R_e=5.2\ \Omega$ , we can compute the current of the circuit by using the Kirchhoffs Voltage Law or KVL:

$\varepsilon=i.R_i+i.R_e=i.(R_i+R_e)$

Solving for i

$\displaystyle i=\frac{\varepsilon}{ R_i+R_e}=\frac{12}{0.06+5.2}=2.28\ A$

i=2.28\ A

The potential difference across the headlight bulbs is

$V=\varepsilon -i.R_i=12\ V-2.28\ A\cdot 0.06\ \Omega=11.86\ V$

$V=11.86\ V$

(b) If the starter motor is operated, taking an additional 35 Amp from the battery, then the total load current is 2.28 A + 35 A = 37.28 A. Thus the output voltage of the battery, that is the voltage that the bulbs have is

$V=\varepsilon -i.R_i=12\ V-37.28\ A\cdot 0.06\ \Omega=9.76\ V$

5 0

3 years ago