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3 years ago

The mass density of an object is 16.3 g/cc, and its volume is determined to be 16.7 cc. what is the mass of the object?

1 answer:

3 0

There are 16.3 grams for every cubic centimeter, and there are 16.7 cubic centimeters. therefore,
16.3 x 16.7

=272.21 g

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Which source is considered a primary source of scientific information? an original research article in a scientific journal a re

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3 years ago

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What is light energy

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2 years ago

A box of mass 26 kg is initially at rest on a flat floor. The coefficient of kinetic friction between the box and the floor is 0

Kazeer [188]

Answer:

$\Delta K = 52J$

Explanation:

The change in kinetic energy will be simply the difference between the final and initial kinetic energies: $\Delta K=K_f-K_i$

We know that the formula for the kinetic energy for an object is:

$K=\frac{mv^2}{2}$

where m is the mass of the object and v its velocity.

For our case then we have:

$\Delta K = K_f-K_i=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}=\frac{m(v_f^2-v_i^2)}{2}$

Which for our values is:

$\Delta K = \frac{m(v_f^2-v_i^2)}{2} = \frac{(26Kg)((2m/s)^2-(0m/s)^2)}{2} = 52J$

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3 years ago

Observe the given figure and find the the gravitational force between m1 and m2.

Leno4ka [110]

Answer:

The gravitational force between m₁ and m₂, is approximately 1.06789 × 10⁻⁶ N

Explanation:

The details of the given masses having gravitational attractive force between them are;

m₁ = 20 kg, r₁ = 10 cm = 0.1 m, m₂ = 50 kg, and r₂ = 15 cm = 0.15 m

The gravitational force between m₁ and m₂ is given by Newton's Law of gravitation as follows;

$F =G \cdot \dfrac{m_{1} \cdot m_{2}}{r^{2}}$

Where;

F = The gravitational force between m₁ and m₂

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

r₂ = 0.1 m + 0.15 m = 0.25 m

Therefore, we have;

$F = 6.67430 \times 10^{-11} \ N \cdot m^2/kg \times \dfrac{20 \ kg\times 50 \ kg}{(0.1 \ m+ 0.15 \ m)^{2}} \approx 1.06789 \times 10^{-6} \ N$

The gravitational force between m₁ and m₂, F ≈ 1.06789 × 10⁻⁶ N

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2 years ago

How far can a sound wave travel in 90 seconds when the ambient air temperature is 10 C?

Ksju [112]

Answer:

s = 30330.7 m = 30.33 km

Explanation:

First we need to calculate the speed of sound at the given temperature. For this purpose we use the following formula:

v = v₀√[T/273 k]

where,

v = speed of sound at given temperature = ?

v₀ = speed of sound at 0°C = 331 m/s

T = Given Temperature = 10°C + 273 = 283 k

Therefore,

v = (331 m/s)√[283 k/273 k]

v = 337 m/s

Now, we use the following formula to calculate the distance traveled by sound:

s = vt

where,

s = distance traveled = ?

t = time taken = 90 s

Therefore,

s = (337 m/s)(90 s)

s = 30330.7 m = 30.33 km

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2 years ago