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3 years ago

Confirm if this is correct or not. If it isn't correct, please correct it.

Physics

1 answer:

kow [346]3 years ago

3 0

Answer:

d = 421.83 m

Explanation:

It is given that,

Height, h = 396.9 m

Horizontal speed, v = 46.87 m/s

We need to find the distance traveled by the ball horizontally. Let t is the time taken by the ball. Using second equation of motion for vertical direction. So,

$396.9=0\times t+\dfrac{1}{2}\times 9.8 t^2\\\\t=9\ s$

Now d is the distance covered by the cannonball. So,

$d=vt\\\\d=46.87\times 9\\\\d=421.83\ m$

Hence, this is the required solution.

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hello friends,i need your help my home work now in physics,topic:motion.40 marks +brainliest if correct .

fenix001 [56]

Answer:

See below

Explanation:

Vertical position is given by

df = do + vo t - 1/2 a t^2 df = final position = 0 (on the ground)

do =original position = 2 m

vo = original VERTICAL velocity = 0

a = acceleration of gravity = 9.81 m/s^2

THIS BECOMES

0 = 2 + 0 * t - 1/2 ( 9.81)t^2

to show t = .639 seconds to hit the ground

During this .639 seconds it flies horizontally at 10 m/s for a distance of

10 m/s * .639 s = 6.39 m

5 0

1 year ago

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Lapatulllka [165]

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4 0

2 years ago

A long, East-West-oriented power cable carrying an

Alla [95]

Answer:

200A

Explanation:

Given that

the distance between earth surface and power cable d = 8m

when the current is flowing through cable , the magnitude flux density at the surface is 15μT

when the current flow throught is zero the magnitude flux density at the surface is 20μT

The change in flux density due to the current flowing in the power cable is

B = 20μT - 15μT

B =5μT -----(1)

The expression of magnitude flux density produced by the current carrying cable is

$B=\frac{\mu_0I}{2\pi d}$ -----(2)

Substitute the value of flux density

B from eqn 1 and eqn 2

$\frac{\mu_0I}{2\pi d}=5\times 10^-^6\\\\\frac{(4\pi \times 10^-^7)I}{2 \pi (8)} =5\times 10^-^6\\\\I=200A$

Therefore, the magnitude of current I is 200A

8 0

2 years ago

A basketball star covers 3.05 m horizontally in a jump to dunk the ball (see figure). His motion through space can be modeled pr

rosijanka [135]

anonymous 4 years agoAny time you are mixing distance and acceleration a good equation to use is ΔY=Viyt+1/2at2 I would split this into two segments - the rise and the fall. For the fall, Vi = 0 since the player is at the peak of his arc and delta-Y is from 1.95 to 0.890. For the upward part of the motion the initial velocity is unknown and the final velocity is zero, but motion is symetrical - it takes the same amount of time to go up as it does to go down. Physiscists often use the trick "I'm going to solve a different problem, that I know will give me the same answer as the one I was actually asked.) So for the first half you could also use Vi = 0 and a downward delta-Y to solve for the time. Add the two times together for the total. The alternative is to calculate the initial and final velocity so that you have more information to work with.

5 0

3 years ago

Read 2 more answers

Two long, straight wires are separated by a distance of 9.15 cm . One wire carries a current of 2.79 A , the other carries a cur

Dafna1 [17]

Answer:

The force is the same

Explanation:

The force per meter exerted between two wires carrying a current is given by the formula

$\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi r}$

where

$\mu_0$ is the vacuum permeability

$I_1$ is the current in the 1st wire

$I_2$ is the current in the 2nd wire

r is the separation between the wires

In this problem

$I_1=2.79 A\\I_2=4.36 A\\r = 9.15 cm = 0.0915 m$

Substituting, we find the force per unit length on the two wires:

$\frac{F}{L}=\frac{(4\pi \cdot 10^{-7})(2.79)(4.36)}{2\pi (0.0915)}=2.66\cdot 10^{-5}N$

However, the formula is the same for the two wires: this means that the force per meter exerted on the two wires is the same.

The same conclusion comes out from Newton's third law of motion, which states that when an object A exerts a force on an object B, then object B exerts an equal and opposite force on object A (action-reaction). If we apply the law to this situation, we see that the force exerted by wire 1 on wire 2 is the same as the force exerted by wire 2 on wire 1 (however the direction is opposite).

3 0

3 years ago