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2 years ago

If a car skids to a stop in 3.8s while undergoing a uniform acceleration of -9.55 m/s2, what is the car’s velocity?

Physics

1 answer:

Lapatulllka [165]2 years ago

3 0

Answer:

36.29 m/s

Explanation:

We know from theory that $v= v_0 +a\Delta t$ . Let's replace the value we know in our equation and solve for the only value left.

$0 = v_0 - 9.55 \cdot 3.8 \\ v_0 = 9.55 \cdot 3.8 = 36.29 m/s$

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b. A string is wrapped around a pulley of radius 0.05 m and moment of inertia 0.2 kg  m2. If the string is pulled with a force

Oduvanchick [21]

Answer:

f = 8 N

Explanation:

Data provided in the question

Radius of the pulley = r = 0.05 m

Moment of inertia = (I) = 0.2 kg.m^{2}

Angular acceleration = ∝ = 2 rad/sec

Based on the above information

As we know that

Torque is

$= force \times radius$

$= f \times r$

And,

Torque is also

$= moment\ of\ inertia \times angular\ acceleration$

$= I \times \alpha$

So,

We can say that

$f \times r = I \times \alpha$

$f \times 0.05 = 0.2 \times 2$

0.05f = 0.4

f = 8 N

We simply applied the above formulas

8 0

3 years ago

You drive a car 640 m to the east, then 340 m to the north what is the magnitude of your displacement

mixer [17]

Magnitude of displacement = $\sqrt{640^2 + 340^2}$

Adding the squares gives displacement = $\sqrt{525,200}$

Displacement = $\sqrt{525,200}$ ≈ 724.7m

6 0

3 years ago

The speed of sound is 346 m/s. If a sound wave travels at a frequency of 55 Hz, what would its wavelength be?

Korvikt [17]

Answer:

6.29 meters.

Explanation:

, where v is the speed of wave and f is the frequency of wave.

We are given that ,

The speed of sound is 346 m/s.

i..e v=346 m/s

A sound wave travels at a frequency of 55 H.

i..e f=55

the wavelength would be 6.29 meters.

This is based on another brainly answer

Link: brainly.com/question/12538018

3 0

2 years ago

Two long parallel wires are separated by forty centimeters and carry oppositely-directed currents of ten amperes. Find the magni

Softa [21]

Answer:

1.04μT

Explanation:

Due to both wires have opposite currents, the magnitude of the total magnetic field is given by

$B_T=\frac{\mu_o I}{2 \pi r_1}-\frac{\mu_o I}{2 \pi r_2}$

I: electric current = 10A

mu_o: magnetic permeability of vacuum = 4pi*10^{-7} N/A^2

r1: distance from wire 1 to the point in which B is measured.

r2: distance from wire 2.

The distance between wires is 40cm = 0.4m. Hence, r1=0.2m r2=0.6m

By replacing in the formula you obtain:

$B_T=\frac{(4\pi *10^{-7}N/A^2)(10A)}{2\pi}(\frac{1}{0.4m}-\frac{1}{0.6m})=1.04*10^{-6}T =1.04\mu T$

hence, the magnitude of the magnetic field is 1.04μT

4 0

3 years ago

A box of paper is labeled 24lb paper. This means that 500 sheets (counted number) of paper size 17in x 22in weighs 24 pounds (Th

Tresset [83]

Answer:

The mass of a single paper is approximately 0.047 lb/paper which in SI Units is approximately 21.77 g/paper

Explanation:

The given information on the size and the weight of paper are;

The mass of a box of 500 sheets of paper = 24 lb

The number of sheets in the paper = 500 sheets

The dimensions of the paper = 17 in. × 22 in., which is equivalent to 43.18 cm × 55.88 cm

The mass of a single paper = The mass of the box of paper/(The number of sheets of paper present in the box)

The mass of a single paper = 24 lb/500 = 0.047 lb/paper

Given that 1 lb = 453.6 g, we have;

0.047 lb/paper = 0.047 lb/paper×453.6 g/(lb) = 21.77 g/paper

The mass of a single paper = 0.047 lb/paper = 21.77 g/paper.

6 0

3 years ago