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3 years ago

A 14n force is applied for 0.33 seconds, calculate the impulse

Physics

1 answer:

Shalnov [3]3 years ago

5 0

Answer:

4.62 N-s

Explanation:

recall that the formula for impulse is given by

Impulse = Force x change in time

in our case, we are given

Force = 14 N

change in time = 0.33s

Simply substituting the above into the equation for impulse, we get

Impulse = Force x change in time

Impulse = 14 x 0.33

= 4.62 N-s

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What will occur if a resistor with high resistance is inserted into a circuit?

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If the resistor is in series with the rest of the circuit then a is the correct answer. The voltage across the resistor in series with another resistor is
$V \frac{R}{R+r}$
where R is the big resistor and r is the small one and V is the total voltage drop across both. This is called a voltage divider

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3 years ago

Read 2 more answers

One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has

Mars2501 [29]

Answer:

6.86 m/s

Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE) = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²

This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

Now, 1/6m(v0)² = mgL ⇒ v0 = $\sqrt{6gL}$

Hence, v0 = 6.86 m/s

4 0

3 years ago

Given a force of 100 N and acceleration of 5 m/s2, what is the mass

tatyana61 [14]

Answer:

20 kg

Explanation:

remember the equation f=ma.

100 N=force

5 m/s2= acceleration

so you need to divide force by acceleration: 100 N/ 5 m/s2= 20 kg, to get the mass.

8 0

2 years ago

Which is the main purpose of an experimental conclusion?

Reptile [31]

Answer:

the Answer is B

because the purpose of experimental conclusion it is to prove rather is a right or wrong is a hypothesis is right or wrong

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3 years ago

An oscillator consists of a block of mass 0.373 kg connected to a spring. When set into oscillation with amplitude 33 cm, the os

aniked [119]

Answer:

(a) T = 0.412s

(b) f = 2.42Hz

(d) k = 86.75N/m

(e) vmax = 5.03 m/s

Explanation:

Given information:

m: mass of the block = 0.373kg

A: amplitude of oscillation = 22cm = 0.22m

T: period of oscillation = 0.412s

(a) The period is the time of one complete oscillation = 0.412s

The period is 0.412s

(b) The frequency is calculated by using the following formula:

$f=\frac{1}{T}=\frac{1}{0.412s}=2.42Hz$

The frequency is 2.42 Hz

(c) The angular frequency is:

$\omega=2\pi f=2\pi (2.42Hz)=15.25\frac{rad}{s}$

The angular frequency is 15.25 rad/s

(d) The spring constant is calculated by solving the following equation for k:

$\omega=\sqrt{\frac{k}{m}}\\\\k=m\omega^2=(0.373kg)(15.25rad/s)^2=86.75\frac{N}{m}$

The spring constant is 86.75N/m

(e) The maximum speed is:

$v_{max}=\omega A=(15.25rad/s)(0.33m)=5.03\frac{m}{s}$

(f) The maximum force applied by the spring if for the maximum elongation, that is, the amplitude:

$F=kA=(86.75N/m)(0.2m)=17.35N$

The maximum force that the spring exerts on the block is 17.35N

3 0

4 years ago