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2 years ago

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You need to measure the lengthof you driveway.which measuring tool would produce accurate results that would expect to be the sa

me as those found by others
A your hands span

B a string cut to the length of the tip of you finger to the bottom of your palm

C a meter stick with only centimeters

D a ruler with millimeters and centimeters

1 answer:

Aleonysh [2.5K]2 years ago

3 0

C a meter stick with only centimeters

D a ruler with millimeters and centimeters

D would be to the nearest half milimeter. Take some time to measure with a 2 inch ruler. Would you really need to know the length to half a mil ?

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A 5.00 g object moving to the right at 20.0 cm/s makes an elastic head-on collision with a 10.0 g object that is initially at re

marusya05 [52]

Answer: a) 6.67cm/s b) 1/2

Explanation:

According to law of conservation of momentum, the momentum of the bodies before collision is equal to the momentum of the bodies after collision. Since the second body was initially at rest this means the initial velocity of the body is "zero".

Let m1 and m2 be the masses of the bodies

u1 and u2 be their velocities respectively

m1 = 5.0g m2 = 10.0g u1 = 20.0cm/s u2 = 0cm/s

Since momentum = mass × velocity

The conservation of momentum of the body will be

m1u1 + m2u2 = (m1+m2)v

Note that the body will move with a common velocity (v) after collision which will serve as the velocity of each object after collision.

5(20) + 10(0) = (5+10)v

100 + 0 = 15v

v = 100/15

v = 6.67cm/s

Therefore the velocity of each object after the collision is 6.67cm/s

b) kinectic energy of the 10.0g object will be 1/2MV²

= 1/2×10×6.67²

= 222.44Joules

kinectic energy of the 5.0g object will be 1/2MV²

= 1/2×5×6.67²

= 222.44Joules

= 111.22Joules

Fraction of the initial kinetic transferred to the 10g object will be

111.22/222.44

= 1/2

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3 years ago

Which statement best describes light, whether polarized or unpolarized?

marshall27 [118]

The best choice would be letter C hope this helps

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3 years ago

Read 2 more answers

A shell is fired at an angle of 35° above the horizontal at a velocity of 40 m/s. (a) What is it's speed at the highest point of

Fudgin [204]

Answer:

Part a)

$v_f = v_x = 32.77 m/s$

Part b)

T = 4.68 s

Explanation:

Part a)

Shell is fired at speed of 40 m/s at angle of 35 degree

so here we have

$v_x = 40 cos35 = 32.77 m/s$

$v_y = 40 sin35 = 22.94 m/s$

since gravity act opposite to vertical speed of the shell so at the highest point of its trajectory the vertical component of the speed will become zero

so at the highest point the speed is given

$v_f = 32.77 m/s$

Part b)

After completing the motion we know that the displacement of the object will be zero in Y direction

so we have

$\Delta y = 0$

$0 = v_y t - \frac{1}{2}gt^2$

$T = \frac{2v_y}{g}$

$T = \frac{2(22.94)}{9.81} = 4.68 s$

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2 years ago

As a quality test of ball bearings, you drop bearings (small metal balls), with zero initial velocity, from a height of 1.94 m i

11Alexandr11 [23.1K]

Answer:

The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

Explanation:

Given that,

Height = 1.94 m

Bounced height = 1.48 m

Time interval $t=14.86\times10^{-3}\ s$

Velocity of the ball bearing just before hitting the steel plate

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2$

Put the value into the formula

$9.8\times1.94=\dfrac{1}{2}\times v^2$

$v=\sqrt{2\times9.8\times1.94}$

$v=6.166\ m/s$

Negative as it is directed downwards

After bounce back,

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2'$

Put the value into the formula

$9.8\times1.48=\dfrac{1}{2}\times v^2'$

$v'=\sqrt{2\times9.8\times1.48}$

$v'=5.38\ m/s$

We need to calculate the average acceleration of the bearings while they are in contact with the plate

Using formula of acceleration

$a=\dfrac{v-v'}{t}$

Put the value into the formula

$a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}$

$a=776.98\ m/s^2$

$a=0.77\times10^{3}\ m/s^2$

Hence,The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

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2 years ago

water vapor contained in a piston–cylinder assembly undergoes an isothermal expansion at 240°c from a pressure of 7 bar to a pre

mafiozo [28]

The ideal gas constant is a proportionality constant that is added to the ideal gas law to account for pressure (P), volume (V), moles of gas (n), and temperature (T) (R). R, the global gas constant, is 8.314 J/K-1 mol-1.

According to the Ideal Gas Law, a gas's pressure, volume, and temperature may all be compared based on its density or mole value.

The Ideal Gas Law has two fundamental formulas.

PV = nRT, PM = dRT.

P = Atmospheric Pressure

V = Liters of Volume

n = Present Gas Mole Number

R = 0.0821atmLmoL K, the Ideal Gas Law Constant.

T = Kelvin-degree temperature

M stands for Molar Mass of the Gas in grams Mol d for Gas Density in gL.

Learn more about Ideal gas law here-

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6 months ago