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Lady_Fox [76]
3 years ago
13

Anna pushes a box with a force of 8.00 newtons. She generates a power of 3.00 watts. How much time does it take for Anna to move

the box the distance of 9.72 meters?
Physics
2 answers:
QveST [7]3 years ago
8 0
Power is the energy in a system per time.  It will have units of Watts which is equal to joules per second. It can be expressed as:

P = E / t

where E = Force x distance

P = Fd / t
t = Fd / P
t = 8 (9.72) / 3.0
t = 25.92 s
Ratling [72]3 years ago
8 0

Answer:

time taken, t = 25.92 seconds

Explanation:

It is given that,

Force acting on the box, F = 8 N

Power generated on the box, P = 3 watts

Distance traveled by the box, d = 9.72 m

We know that the power generated by an object is equal to the total work done divided by total time taken. Its formula is given by :

P=\dfrac{W}{t}

Since, W = F d

P=\dfrac{Fd}{t}

t=\dfrac{Fd}{P}

t=\dfrac{8\ N\times 9.72\ m}{3\ W}

t = 25.92 seconds

So, the time taken by Anna to move the box to the distance of 9.72 meters is 25.92 seconds. Hence, this is the required solution.

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A tiger is running across a field. What would happen if the hydrogen ion pumps into the tigers mitochondria stopped functioning
galben [10]

Answer:

The tiger would not be able to produce glucose causing it to stop running

Explanation:

Since the mitochondria is in charge of producing ATP the tiger would not be able to use any glucose causing it to not be able to run.

3 0
3 years ago
What would most likely cause the future acceleration of the expansion of the universe
irga5000 [103]

Answer:

never scared

that might help you

3 0
3 years ago
(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th
julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by

V = -\dfrac{kQ}{r}

where k = 9\times 10^9 \text{ F/m}

Difference in potential between the points is

kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}

PD = 135\times 10^3\text{ V} = 135\text{ kV}

(b)

If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.

270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]

10 - \dfrac{1}{x} = \dfrac{270000}{9\times10^9\times3\times10^{-6}} = 10

\dfrac{1}{x} = 0

x = \infty

The charge chould be moved to infinity

7 0
3 years ago
you have been called to testify as a as an expert witness in a trial involving a head-on collision Car A weighs 1515 pounds and
GrogVix [38]

Answer:

70.6 mph

Explanation:

Car A mass= 1515 lb

Car B mass=1125 lb  

Speed of car B is 46 miles/h

Distance before locking, d=19.5 ft

Coefficient of kinetic friction is 0.75

Initial momentum of car B=mv where m is mass and v is velocity in ft/s  

46 mph*1.46667=67.4666668 ft/s

Momentum_B=1125*67.4666668 ft/s

Initial momentum of car A is given by

Momentum_A=1515v_a where v_a is velocity of A

Taking East as positive and west as negative then the sum of initial momentum is

1515v_a-(1125*67.4666668 ft/s)

The common velocity is represented as v_c hence after collision, the final momentum is

Momentum_final=(m_a+m_b)v_c=(1515+1125)v_c=2640v_c

From the law of conservation of linear momentum, sum of initial and final momentum equals each other hence

1515v_a-(1125*67.4666668 ft/s)= 2640v_c

The acceleration of two cars a=-\mug=-0.75*32.17=-24.1275 ft/s^{2}

From kinematic equation

v^{2}=u^{2}+2as hence

v^{2}-u^{2}=2as

0^{2}-(v_c)^{2}=2*-24.1275*19.5

v_c=\sqrt{2*24.1275*19.5}=30.67 ft/s

Substituting the value of v_c in equation 1515v_a-(1125*67.4666668 ft/s)= 2640v_c

1515v_a-(1125*67.4666668 ft/s)= 2640*30.67

1515v_a=(1125*67.4666668 ft/s)+2640*30.67

v_a=\frac {156868.8}{1515}=103.5438 ft/s

\frac {103.5438}{1.46667}=70.59787 mph\approx 70.60 mph

3 0
3 years ago
Can someone give me an example or explanation of calculating energy from voltage? Many thanks!
antoniya [11.8K]
Expression to calculate energy from voltage: E= V*Q where E= energy, V= voltage, and Q= charge

Additional help:
-To find the Voltage ( V )
[ V = I x R ] V (volts) = I (amps) x R (Ω)

-To find the Current ( I )
[ I = V ÷ R ] I (amps) = V (volts) ÷ R (Ω)

-To find the Resistance ( R )
[ R = V ÷ I ] R (Ω) = V (volts) ÷ I (amps)

I hope that helps to some extent-
7 0
3 years ago
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