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3 years ago

12

A soccer ball initially at rest rolls down a hill with an acceleration of 3.0 m /s2 .it accelerates for 5 seconds before melissa

is able to catch it how far did it roll ?

1 answer:

slavikrds [6]3 years ago

6 0

The correct answer is the Soccer ball is 3.2 plus 5 that’s your answer.

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HELLLLLP PLEASE, thank u :)

MrRa [10]

Answer:

Option B. 3.0×10¯¹¹ F.

Explanation:

The following data were obtained from the question:

Potential difference (V) = 100 V.

Charge (Q) = 3.0×10¯⁹ C.

Capacitance (C) =..?

The capacitance, C of a capacitor is simply defined as the ratio of charge, Q on either plates to the potential difference, V between them. Mathematically, it is expressed as:

Capacitance (C) = Charge (Q) / Potential difference (V)

C = Q/V

With the above formula, we can obtain the capacitance of the parallel plate capacitor as follow:

Potential difference (V) = 100 V.

Charge (Q) = 3.0×10¯⁹ C.

Capacitance (C) =..?

C = Q/V

C = 3.0×10¯⁹ / 100

C = 3.0×10¯¹¹ F.

Therefore, the capacitance of the parallel plate capacitor is 3.0×10¯¹¹ F.

7 0

3 years ago

51. Suppose you measure the terminal voltage of a 3.200-V lithium cell having an internal resistance of 5.00Ω by placing a 1.00-

Tanya [424]

Explanation:

Given that,

Terminal voltage = 3.200 V

Internal resistance $r= 5.00\ \Omega$

(a). We need to calculate the current

Using rule of loop

$E-IR-Ir=0$

$I=\dfrac{E}{R+r}$

Where, E = emf

R = resistance

r = internal resistance

Put the value into the formula

$I=\dfrac{3.200}{1.00\times10^{3}+5.00}$

$I=3.184\times10^{-3}\ A$

(b). We need to calculate the terminal voltage

Using formula of terminal voltage

$V=E-Ir$

Where, V = terminal voltage

I = current

r = internal resistance

Put the value into the formula

$V=3.200-3.184\times10^{-3}\times5.00$

$V=3.18\ V$

(c). We need to calculate the ratio of the terminal voltage of voltmeter equal to emf

$\dfrac{Terminal\ voltage}{emf}=\dfrac{3.18}{3.200 }$

$\dfrac{Terminal\ voltage}{emf}= \dfrac{159}{160}$

Hence, This is the required solution.

5 0

3 years ago

Review. A light spring of force constant 3.85 N/m is compressed by 8.00 cm and held between a 0.250-kg block on the left and a 0

photoshop1234 [79]

The answer is 0.245N.

<h3>What is kinetic energy?</h3>

A particle or an item that is in motion has a sort of energy called kinetic energy. An item accumulates kinetic energy when work, which involves the transfer of energy, is done on it by exerting a net force.
Kinetic energy comes in five forms: radiant, thermal, acoustic, electrical, and mechanical.
The energy of a body in motion, or kinetic energy (KE), is essentially the energy of all moving objects. Along with potential energy, which is the stored energy present in objects at rest, it is one of the two primary types of energy.
Explain that a moving object's mass and speed are two factors that impact the amount of kinetic energy it will possess.

(b) 0.100

For the block on the left, $f_{k} =u_{k} n= 0.100(2.45N)=0.245N.$

∑ $F_{x}$ = $ma_{x}$

–0.308N+0.245N=(0.250kg)a

a=−0.252 $m/s^{2}$ if the force of static friction is not too large.

For the block on the right, $f_{k}$ = $u_{k} n$ =0.490N. The maximum force of static friction would be larger, so no motion would begin, and the acceleration is zero

To learn more about kinetic energy, refer to:

brainly.com/question/25959744

#SPJ4

7 0

2 years ago

A 125kg bumper car going 18.5 m/a bumps a 187.kkg bumper car at rest. if the first car (125kg) bounces back at 8 m/s what is the

kozerog [31]

<span><span>Imagine we have a 2 lb ball of putty moving with a speed of 5 mph striking and sticking to a 18 lb bowling ball at rest; the time it takes to collide is 0.1 s. After the collision, the two move together with a speed of v1. To find v1, use momentum conservation: 2x5=(18+2)v1, v1=0.5 mph. </span><span>Next, imagine we have a 18 lb bowling ball moving with a speed of 5 mph striking and sticking to a 2 lb ball of putty at rest; the time it takes to collide is 0.1 s. After the collision, the two move together with a speed of v2. To find v2, use momentum conservation: 18x5=(18+2)v2, v2=4.5 mph. </span><span>
</span><span>
</span><span>now figure out your problem its really easy let me know if you need more help </span></span>

3 0

3 years ago

A 2.60 kg lion runs at a speed of 5.00 m/s until he sees his prey. The lion then speeds up 8.00 m/s to catch it. How much work d

ELEN [110]

Answer: 50.7 J

Explanation:

Given

mass of lion is $m=2.6\ kg$

The initial speed of the lion is $v_1=5\ m/s$

increased speed of lion is $v_2=8\ m/s$

Initially, its kinetic energy is $K_1=\frac{1}{2}mv_1^2$

Final kinetic energy $k_2=\frac{1}{2}mv_2^2$

work did by lion after speed up is $k_2-k_1$

$\Rightarrow W=\dfrac{1}{2}\times 2.6[8^2-5^2]\\\\\Rightarrow W=1.3\times [39]=50.7\ J$

5 0

2 years ago