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sp2606 [1]
2 years ago
7

The world record for pole vaulting is 6.15 m. If the pole vaulter's gravitational potential energy is 4942 J, what is his mass?

g 9.8 m/s2
Physics
1 answer:
navik [9.2K]2 years ago
7 0
The gravitational potential energy is calculated by multiplying the mass of the object to the height and the gravitational acceleration which is 9.8 m/s^2. We do as follows:

GPE = mgh
GPE = 4942
4942 = m (9.8)(6.15)
m = 82 kg 

Hope this helps.
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In a certain region of space, a uniform electric field has a magnitude of 4.30 x 104 n/c and points in the positive x direction.
denis23 [38]
The magnetic force exerted by a field E to a charge q is given by F=Eq. In this case, F=4.30*10^4*(6.80mu C). 1mu C=10^-6C, so F=4.30*6.80=10^-2=0.29N. The direction is in the x direction, the direction that the field is applied because the charge is positive.
5 0
2 years ago
Two separate but nearby coils are mounted along the same axis. A power supply controls the flow of current in the first coil, an
gulaghasi [49]

Complete Question

The complete question is shown on the first uploaded image

Answer:

  The correct answer is option c

Explanation:

Faraday states that when there is a change in magnetic field of a coil of a wire, it means that there exist an emf in the circuit which in induced due to the change in the magnetic flux

From the question  two separate but nearby coils are mounted along the axis. First coil is connected to the power supply and the current flow is controlled by the supply.When the current alternates, it would produce magnetic field ,also the second coil is connected to an ammeter which indicates the current that is flowing in it when current in the first coil changes

This magnetic field that is produce would cause a change flux which would induce current in the second coil so the ammeter would indicate current flow in the second coil

a is incorrect because the current in fir coil is not change hence flux won't change therefore current is is not induced in second coil

This is the same reason b is incorrect

d is incorrect due to the fact that when the second coil is connected to a power supply by rewiring it to be in series with first coil the law of electromagnetism would no longer hold so he ammeter would show no reading  

6 0
2 years ago
A truck with a heavy load has a total mass of 7100 kg. It is climbing a 15∘ incline at a steady 15 m/s when, unfortunately, the
Andrej [43]

Answer:

The load has a mass of 2636.8 kg

Explanation:

Step 1 : Data given

Mass of the truck = 7100 kg

Angle = 15°

velocity = 15m/s

Acceleration = 1.5 m/s²

Mass of truck = m1 kg

Mass of load = m2 kg

Thrust from engine = T

Step 2:

⇒ Before the load falls off, thrust (T) balances the component of total weight downhill:

T = (m1+m2)*g*sinθ

⇒ After the load falls off, thrust (T) remains the same but downhill component of weight becomes  m1*gsinθ .

Resultant force on truck is F = T – m1*gsinθ  

F causes the acceleration of the truck: F= m*a

This gives the equation:

T – m1*gsinθ = m1*a  

T = m1(a + gsinθ)

Combining both equations gives:

(m1+m2)*g*sinθ = m1*(a + gsinθ)

m1*g*sinθ + m2*g*sinθ =m1*a + m1*g*sinθ

m2*g*sinθ = m1*a

Since m1+m2 = 7100kg, m1= 7100 – m2. This we can plug into the previous equation:

m2*g*sinθ = (7100 – m2)*a

m2*g*sinθ = 7100a – m2a

m2*gsinθ + m2*a = 7100a

m2* (gsinθ + a) = 7100a

m2 = 7100a/(gsinθ  + a)

m2 = (7100 * 1.5) / (9.8sin(15°) + 1.5)

m2 = 2636.8 kg

The load has a mass of 2636.8 kg

6 0
2 years ago
A boy uses a slingshot to launch a pebble straight up into the air. The pebble reaches a height of 37.0 m above the launch point
denis-greek [22]

Answer:

V0=27.4 m/s; t=0.8 s

Explanation:

Final position y=37.0 m, time = 2.3 s; Initial position is set to be zero. We calculate the initial speed with the kinematics equation:

y_f=v_0t-0.5*g*t^2 We solve for initial speed

v_0=\frac{y_f+0.5gt^2}{t}=\frac{37+4.9*2.3^2}{2.3}=27.4m/s

Now, using the same expression we estimated time to first reach 18.5 m :

18.5=27.4t-4.9t^2 Second order equation with solutions

t1=0.8 s and t2=4.8 s

The first time corresponds to the first reach.

7 0
2 years ago
A very long uniform line of charge has charge per unit length λ1 = 4.80 μC/m and lies along the x-axis. A second long uniform li
Elodia [21]

Answer:

a) E=228391.8 N/C

b) E=-59345.91N/C

Explanation:

You can use Gauss law to find the net electric field produced by both line of charges.

\int \vec{E_1}\cdot d\vec{r}=\frac{\lambda_1}{\epsilon_o}\\\\E_1(2\pi r)=\frac{\lambda_1}{\epsilon_o}\\\\E_1=\frac{\lambda_1}{2\pi \epsilon_o r_1}\\\\\int \vec{E_2}\cdot d\vec{r}=\frac{\lambda_2}{\epsilon_o}\\\\E_2=\frac{\lambda_2}{2\pi \epsilon_o r_2}

Where E1 and E2 are the electric field generated at a distance of r1 and r2 respectively from the line of charges.

The net electric field at point r will be:

E=E_1+E_2=\frac{1}{2\pi \epsilon_o}(\frac{\lambda_1}{r_1}+\frac{\lambda_2}{r_2})

a) for y=0.200m, r1=0.200m and r2=0.200m:

E=\frac{1}{2\pi(8.85*10^{-12}C^2/Nm^2)}[\frac{4.80*10^{-6}C}{0.200m}-\frac{2.26*10^{-6}C}{0.200m}}]=228391.8N/C

b) for y=0.600m, r1=0.600m, r2=0.200m:

E=\frac{1}{2\pi(8.85*10^{-12}C^2/Nm^2)}[\frac{4.80*10^{-6}C}{0.600m}-\frac{2.26*10^{-6}C}{0.200m}}]=-59345.91N/C

5 0
2 years ago
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