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2 years ago

I need help with those 2.

Physics

1 answer:

Nimfa-mama [501]2 years ago

5 0

Hi there!

Question 1:

We know that Impulse = Δp = mΔv, so:

I = 0.045(70) = 3.15 Ns

I = F · t, so:

3.15/0.1 = F

F = 31.5 N

Question 2.

We can find the impulse using:

I = Ft

I = 300 · 0.04 = 12 Ns

Find the change in velocity using:

I = mΔv

12/0.13 = Δv = 92.31 m/s

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Work occurs when

defon

I believe the answer is the second option.

7 0

3 years ago

The overall length of a piccolo is 32.0 cm. The resonating air column vibrates as in a pipe that is open at both ends. (a) Find

lana66690 [7]

Answer:

lowest frequency = 535.93 Hz

distance between adjacent anti nodes is 4.25 cm

Explanation:

given data

length L = 32 cm = 0.32 m

to find out

frequency and distance between adjacent anti nodes

solution

we consider here speed of sound through air at room temperature 20 degree is approximately v = 343 m/s

lowest frequency will be = $\frac{v}{2L}$ ..............1

put here value in equation 1

lowest frequency will be = $\frac{343}{2(0.32)}$

lowest frequency = 535.93 Hz

and

we have given highest frequency f = 4000Hz

wavelength = $\frac{v}{f}$ ..............2

put here value

wavelength = $\frac{343}{4000}$

wavelength = 0.08575 m

so distance = $\frac{wavelength}{2}$ ..............3

distance = $\frac{0.08575}{2}$

distance = 0.0425 m

so distance between adjacent anti nodes is 4.25 cm

3 0

3 years ago

how does inertia explain the movement of your body when a car in which you are riding comes to a sudden stop? (8 sentences)

nydimaria [60]

Your body continues to move unless stopped by the seatbelt. An object in motion will remain in motion. Since your body was already moving it will continue to.

5 0

3 years ago

An electron with a speed of 1.2 × 107 m/s moves horizontally into a region where a constant vertical force of 5.2 × 10-16 N acts

Aliun [14]

Answer: 0.642mm

Explanation: F= force = 5.2×10^-16 N,

v = velocity of electron = 1.2×10^7 m/s,

m = mass of electron = 9.11×10^-31 kg.

We will assume the motion of the object to be of a constant acceleration, hence newton's laws of motion is applicable.

Recall that f = ma.

Where a = acceleration

This acceleration of vertical because it occurred when the object deflected.

5.2×10^-16 = 9.11×10^-31 (ay)

ay = 5.2×10^-16 / 9.11×10^-31

ay = 5.71×10^14 m/s²

For the horizontal motion, x = vt

Where x = horizontal distance = 0.019m and v is the velocity = 1.2×10^7 m/s,

By substituting the parameters, we have that

0.019 = 1.27×10^7 × t

t = 0.019 / 1.27 × 10^7

t = 1.5×10^-9 s

The vertical distance (y) is gotten by using the formulae below

y = ut + at²/2

but u = 0

y = at²/2

y = 5.71×10^14 × (1.5×10^-9)²/2

y = 0.00128475/2

y = 0.000642m = 0.642mm

7 0

3 years ago

Read 2 more answers

A 5.00 kg object oscillates back and forth at the end ofa spring whose spring constant is 49.3 N/m. An obersever istraveling at

defon

Answer:

5.571 sec

Explanation:

angular frequency = √ (k/m) = √ (49.3 / 5) = 3.14 rad/s

Period To = 2π / angular frequency

Period To = 2π/3.14 = 2 × 3.14 / 3.142 = 2.00 sec which you got

T measured by the observer = To / (√ (1 - (v²/c²))) = 2 / √( 1 - 0.871111) = 2 / 0.35901 = 5.571 sec

t=2.00/(1-√((2.80*10^8)^2/(3.00*10^8)^2))= should have been ( To / (√ (1 - (v²/c²))). where To = 2.00 sec

8 0

3 years ago