3 years ago

What is the average velocity of the time interval 0 to 4 seconds

1 answer:

7 0

Average velocity over a given time interval is the distance traveled divided by the time:
$v= \frac{60m-20m}{4s-0s} =40/4=10 \frac{m}{s}$

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A chunk of paraffin (wax) has a mass of 50.4 grams and a volume of 57.9 cm3. What is the density of

ExtremeBDS [4]

$Formula\ for\ density:\\\\
p=\frac{m}{V}\\p-density,\\m-mass,\\V-volume\\\\
Data:\\
m=50,4grams\\
V=57,9cm^3\\\\
p=\frac{50,4g}{57,9cm^3}=0,87\frac{g}{cm^3}\\\\Density\ of\ paraffin\ is\ equal\ to\ 0,87\frac{g}{cm^3}.
$

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3 years ago

a block measures 3.5 cm long 2.8 cm wide and 1.6 cm deep. the density of the block 2.5 g/cm. calculate the volume of the block.

luda_lava [24]

Volume of a block can be found by: length × width × height. So:

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2 years ago

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A 75.0-kg ice skater moving at 10.0 m/s crashes into a stationary skater of equal mass. After the collision, the two skaters mov

Ksenya-84 [330]

Answer:

Explanation:

Momentum change for either skater is mΔv = 75.0(5.0) = 375 kg•m/s

As a change in momentum is equal to an impulse

375 = FΔt

F = 375/0.100 = 3750 N

As 3750 N < 4500 N no bones are broken.

4 0

2 years ago

A ball is dropped from rest. how fast is the ball going after 3 seconds

Airida [17]

Answer:

v = 29.4m/s

Explanation:

Since the ball is dropped at rest,

u = 0m/s

a = 9.81m/s²

Using

v = u + at

After 3 seconds,

v = 0 + (9.81)(3)

v = 29.4m/s

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2 years ago

A 4.0 m length of gold wire is connected to a 1.5 V battery, and a current of 4.0 mA flows through it. What is the diameter of t

sladkih [1.3K]

Explanation:

Given that,

Length of gold wire, l = 4 m

Voltage of battery, V = 1.5 V

Current, I = 4 mA

The resistivity of gold, $\rho=2.44\times 10^{-8}\ \Omega-m$

Resistance in terms of resistivity is given by :

$R=\dfrac{\rho l}{A}$

Also, V = IR

So,

$\dfrac{V}{I}=\dfrac{\rho l}{A}$

A is area of wire,

$\dfrac{V}{I}=\dfrac{\rho l}{\pi r^2}$ , r is radius, r = d/2 (diameter=d)

$\dfrac{V}{I}=\dfrac{\rho l}{\pi (d/2)^2}\\\\\dfrac{V}{I}=\dfrac{4\rho l}{\pi d^2}\\\\d=\sqrt{\dfrac{4\rho l I}{V\pi}} \\\\d=\sqrt{\dfrac{4\times 2.44\times 10^{-8}\times 4\times 4\times 10^{-3}}{1.5\times \pi}} \\\\d=18.2\ \mu m$

Out of four option, near option is (C) 17 μm.

6 0

2 years ago