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1 year ago

Accelerates uniformly at 2.0 ms2 for 10.0s. Calculate its final velocity

Physics

1 answer:

Crazy boy [7]1 year ago

4 0

Answer:

The distance is

Explanation:

Apply the equation of motion

(

)

The initial velocity is

−

The acceleration is

−

Therefore, when

, we get

(

)

⋅

and when

(

)

⋅

Therefore,

The distance travelled in the fourth second is

(

)

−

(

)

−

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A rifle of mass 2 kg is horizontally suspended by a pair of strings so that recoil can be measured. The rifle fires a bullet of

aliya0001 [1]

Answer: 1m/s

Explanation: according to the law of conservation of linear momentum in an isolated system, the momentum of the gun equals that of the bullet.

Mathematically

Mb×Vb = Mg×Vg

Where Mb = mass of bullet = 1/100 = 0.01 kg

Vb = velocity of bullet = 200 m/s

Mg = mass of gun = 2kg

Vg = recoil velocity of gun =?

0.01×200 = 2×Vg

Vg = 0.01×200/2

Vg = 0.01×100

Vg = 1m/s

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2 years ago

A disk of radius R = 7.52 cm is centered at the origin and lies along the y – z plane. The disk has a surface charge density σ =

mr Goodwill [35]

Answer:

The electric field produced by this disk along the x axis at point (P = 1.01 m, 0.00 m) is 996.54 N/C

Explanation:

The electric field produced by this disk along the x axis at point (P = 1.01 m, 0.00 m), will be evaluated as follows:

Since x > 0

$E_{x} = 2\pi \sigma k [1-\frac{x}{\sqrt{x^2 +R^2}}]$

σ is surface charge density = 5.88 × 10⁻⁶ C / m²

R is the radius = 7.52cm = 0.0752m

position x = 1.01m

k is coulomb force constant = 8.99 × 10⁹ Nm² / C²

$E_{x} = 2\pi ( 5.88X10^{-6})(8.99 X10^9)[1-\frac{1.01}{\sqrt{1.01^2 +0.0752^2}}]$

= 996.54 N/C

Therefore, The electric field produced by this disk along the x axis at point (P = 1.01 m, 0.00 m) is 996.54 N/C

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3 years ago

Please Help Quick ASAP Hurry This is Physical Science

krok68 [10]

Answer:

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Explanation:

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2 years ago

Which state of matter has a definite volume but no definite shape

a_sh-v [17]

The answer is liquid.

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3 years ago

What current is needed to generate the magnetic field strength of 5.0×10−5T at a point 1.5 cm from a long, straight wire? Expres

mixer [17]

Answer:

3.7 A

Explanation:

Parameters given:

Magnetic field strength, B = 5 * 10^(-5) T

Distance of magnetic field from wire, r = 1.5 cm = 0.015 m

The magnetic field, B, due to a current, I, flowing a wire is given as:

B = (μ₀*I) / 2πr

Where μ₀ = permeability of free space

To get the current, I, we make I the subject of the formula:

I = (2πr * B) / μ₀

I = (2 * 3.142 * 5 * 10^(-5)) / (1.25663706 × 10^(-6))

I = 3.7 A

4 0

3 years ago

Accelerates uniformly at 2.0 ms2 for 10.0s. Calculate its final velocity​

Accelerates uniformly at 2.0 ms2 for 10.0s. Calculate its final velocity