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3 years ago

Knowing that 1 inch is equal to 2.54 centimeters, convert 287 cm to inches.

Physics

2 answers:

umka2103 [35]3 years ago

6 0

Answer: 112.992

Explanation: If you divide 287 by 2.54, you will get 112.992125984. I just got rid of the five numbers after 992.

Misha Larkins [42]3 years ago

5 0

1 inch = 2.54 cm

?= 287 cm let, ? be (a)

cross multiply

2.54a= 1×287

a= 287/2.54

a= 112.992126

therefore 287 cm = 112.992126 inches.

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Soil formation involves both

Softa [21]

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6 0

3 years ago

Christina drives his moped 7 kilometers North. She stops for lunch and then drives 5 kilometers East. What distance did she cove

Mashutka [201]

Answer:

She covers the distance is 12 km.

The magnitude of displacement is 8.6 km.

The direction of her displacement is north east.

Explanation:

Given that,

Christina drives his moped 7 kilometers North and stop for lunch and then drive 5 km east.

We need to calculate the total distance

Using formula of distance

$d=d_{1}+d_{2}$

Put the value into the formula

$d=7+5$

$d=12\ km$

We need to calculate the magnitude of displacement

Using formula of displacement

$D=xi+yj$

$D=5i+7j$

$D=\sqrt{5^2+7^2}$

$D= 8.6\ km$

The direction of her displacement is north east.

Hence, She covers the distance is 12 km.

The magnitude of displacement is 8.6 km.

The direction of her displacement is north east.

6 0

3 years ago

Projectile's horizontal range on level ground is R=v20sin2θ/g. At what launch angle or angles will the projectile land at half o

seraphim [82]

Answer:

$\theta = 15^o \: or\: 75^o$

Explanation:

As we know that the formula of range is given as

$R = \frac{v^2sin2\theta}{g}$

now we know that

maximum value of the range of the projectile is given as

$R_{max} = \frac{v^2}{g}$

now we need to find such angles for which the range is half the maximum value

so we will have

$\frac{R}{2} = \frac{v^2}{2g} = \frac{v^2sin(2\theta)}{g}$

$sin(2\theta) = \frac{1}{2}$

$2\theta = 30 or 150$

$\theta = 15^o \: or\: 75^o$

7 0

3 years ago

You are designing a flywheel. It is to start from rest and then rotate with a constant angular acceleration of 0.200 rev/s^2. Th

Rama09 [41]

Answer:

I = 8.75 kg m

Explanation:

This is a rotational movement exercise, let's start with kinetic energy

K = ½ I w²

They tell us that K = 330 J, let's find the angular velocity with kinematics

w² = w₀² + 2 α θ

as part of rest w₀ = 0

w = √ 2α θ

let's reduce the revolutions to the SI system

θ = 30.0 rev (2π rad / 1 rev) = 60π rad

let's calculate the angular velocity

w = √(2 0.200 60π)

w = 8.683 rad / s

we clear from the first equation

I = 2K / w²

let's calculate

I = 2 330 / 8,683²

I = 8.75 kg m

4 0

3 years ago

a) What is the average useful power output of a person who does6.00×10^6Jof useful work in 8.00 h? (b) Working at this rate, how

NARA [144]

Answer:

208.33 W

141.26626 seconds

Explanation:

E = Energy = $6\times 10^6\ J$

t = Time taken = 8 h

m = Mass = 2000 kg

g = Acceleration due to gravity = 9.81 m/s²

h = Height of platform = 1.5 m

Power is obtained when we divide energy by time

$P=\frac{E}{t}\\\Rightarrow P=\frac{6\times 10^6}{8\times 60\times 60}\\\Rightarrow P=208.33\ W$

The average useful power output of the person is 208.33 W

The energy in the next part would be the potential energy

The time taken would be

$t=\frac{E}{P}\\\Rightarrow t=\frac{mgh}{208.33}\\\Rightarrow t=\frac{2000\times 9.81\times 1.5}{208.33}\\\Rightarrow t=141.26626\ s$

The time taken to lift the load is 141.26626 seconds

5 0

4 years ago