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3 years ago

A standard 1 kilogram weight is a cylinder 41.5 mm in height and 44.0 mm in diameter. what is the density of the material?

Physics

1 answer:

Korvikt [17]3 years ago

6 0

Answer;

=15855.40 kg/m^3

Explanation;

Volume (V) of the cylinder = pi x r^2 x h

V = 3.14 x (44/2 x 10^-3)^2 x 41.5 x 10^-3

V = 6.307 x 10^-5 m^3

By density = m/V

mass = 1 kg

density = 1/(6.307 x 10^-5) = 15855.40 kg/m^3

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What are used for manufacturing paper along with chemical

Airida [17]

Answer:

Dry-strength additives, or dry-strengthening agents, are chemicals that improve paper strength normal conditions. These improve the paper's compression strength, bursting strength, tensile breaking strength, and delamination resistance. Typical chemicals used include cationic starch and polyacrylamide derivatives.

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2 years ago

BRAINLEST FOR CORRECT ANSWER

xxMikexx [17]

Answer:

3kg sledgehammer swung at 1.5 m/s

Explanation:

Small Sledgehammer:

Mass:3.0

Velocity:1.5

MASS×VELOCITY=MOMENTUM

3.0×1.5= 4.5 (momentum)

Large Sledgehammer:

Mass:4.0

Velocity:0.9

4.0×0.9=3.6 (momentum)

higher momentum is the smaller Sledgehammer.

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3 years ago

How much electrical energy is used by a 400 W toaster that is operating for 5 minutes?

Shkiper50 [21]

The electrical energy consumed by a toaster is 0.033 Kwh.

<u>Explanation</u>:

The power utilized by the toaster is 400 W.

$\text { The power utilized by the toaster is } 400 \mathrm{W} \text { in kilo-watts is } \frac{400}{1000}=0.4 \mathrm{Kw}$

The toaster is operated for 5 Minutes.

$\text { The toaster is operated for } 5 \text { Minutes in hours is } \frac{5}{60}=0.083 \text { hours. (One minute is } 60 \text { seconds) }$

We know that,

$\text {power}=\frac{\text {energy}}{\text {time}}$

Substitute the values in the above formula to obtain electrical energy,

$0.4=\frac{\text { energy }}{0.083}$

Electrical energy = 0.4 × 0.083

Electrical energy is 0.033 Kwh.

3 0

3 years ago

A hockey puck slides off the edge of a platform with an initial velocity of 20 m/s horizontally. The height of the platform abov

Rina8888 [55]

Answer:

20.96 m/s

Explanation:

Using the equations of motion

y = uᵧt + gt²/2

Since the puck slides off horizontally,

uᵧ = vertical component of the initial velocity of the puck = 0 m/s

y = vertical height of the platform = 2 m

g = 9.8 m/s²

t = time of flight of the puck = ?

2 = (0)(t) + 9.8 t²/2

4.9t² = 2

t = 0.639 s

For the horizontal component of the motion

x = uₓt + gt²/2

x = horizontal distance covered by the puck

uₓ = horizontal component of the initial velocity = 20 m/s

g = 0 m/s² as there's no acceleration component in the x-direction

t = 0.639 s

x = (20 × 0.639) + (0 × 0.639²/2) = 12.78 m

For the final velocity, we'll calculate the horizontal and vertical components

vₓ² = uₓ² + 2gx

g = 0 m/s²

vₓ = uₓ = 20 m/s

Vertical component

vᵧ² = uᵧ² + 2gy

vᵧ² = 0 + 2×9.8×2

vᵧ = 6.26 m/s

vₓ = 20 m/s, vᵧ = 6.26 m/s

Magnitude of the velocity = √(20² + 6.26²) = 20.96 m/s

4 0

3 years ago

Read 2 more answers

A model rocket accelerates at 15.3 m/s2 with a force of 44 N.

melamori03 [73]

To calculate the mass of the body moving, we use Newton's second law of motion which is F = ma where F is the force, m is the mass of the object and a is its acceleration.

F = ma
44 = m(15.3)
m = 2.9 kg

The mass of the rocket would be 2.9 kg.

8 0

2 years ago

Read 2 more answers