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3 years ago

6

A horizontal force of 350N is exerted on a 2.5 kg ball as it rotates uniformly in a horizontal circle of radius of 0.90m. Calcul

ate the speed of the ball.

1 answer:

harkovskaia [24]3 years ago

5 0

F=mv^2/R
----> V^2=FR/m=(350x0.9)/2.5=126
----- V=11.22 m/s

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Circle the statements that indicate a chemical property.

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C) because a new substance is formed
B) not sure but might be because the chemical properties of the substance has changed

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2 years ago

two identical-looking, large, round balls are placed in front of you. one is filled with feathers and the other is filled with s

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If you apply a little bit of force, one will move easier than the other since it is lighter.

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2 years ago

Kids are pelting a window with snowballs. On average, two snowballs of roughly 300-g mass hit the window each second, moving hor

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Answer:

The average force exerted on the window due to two snowballs is 6 N

Explanation:

Given:

Mass of snowballs $m = 300 \times 10^{-3}$ Kg

Velocity of snowball $v = 10 \frac{m}{s}$

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$F = \frac{dP}{dt}$

Here, final velocity is zero so we write,

$F = \frac{mv}{1}$

Where $dt = 1$ sec

$F = 300 \times 10^{-30} \times 10$

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$F = 3 \times 2$

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Therefore, the average force exerted on the window due to two snowballs is 6 N

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3 years ago

A long, straight, horizontal wire carries a left-to-right current of 40 A. If the wire is placed in a uniform magnetic field of

Drupady [299]

Answer:

$4.5\times 10^{-5} T$

Explanation:

We are given that

Current in wire=40 A

Magnetic field= $B_1=3.5\times 10^{-5}$ T( vertically downward)

We have to find the resultant magnitude of the magnetic field 29 cm above the wire and 29 cm below the wire.

According to Bio-Savart law, the magnetic field exerted by the wire at distance R is given by

$B_{wire}=B_2=\frac{\mu_0I}{2\pi R}$

We have R=29 cm= $\frac{29}{100}=0.29 m$

1 m=100 cm

Substitute the values in the given formula

$B_2=\frac{4\pi\times 10^{-7}\times 40}{2\times \pi\times 0.29}=\frac{2\times 40\times 10^{-7}}{0.29}=2.76\times 10^{-5} T$

The resultant magnetic field is given by

$B=\sqrt{B^2_1+B^2_2}$

Substitute the values then we get

$B=\sqrt{(3.5\times 10^{-5})^2+(2.76\times 10^{-5})^2}$

$B=4.5\times 10^{-5} T$

The resultant magnitude of magnetic field is same above and below the wire as it is at same distance.

The resultant magnitude of the magnetic field 29 cm below the wire= $4.5\times 10^{-5} T$

Hence, the resultant magnitude of the magnetic field 29 cm above the wire= $4.5\times 10^{-5} T$

7 0

2 years ago

A 100 kg bungee jumper leaps from a bridge. The bungee cord has an un-streched equilibrium length of 10 m, and a spring constant

nika2105 [10]

Answer:

11.78meters

Explanation:

Given data

Mass m = 100kg

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That is

Us=Ug

Us= 1/2kx^2

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Hence the bungee jumper will reach 1.78+10= 11.78meters below the surface of the bridge

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2 years ago

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