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3 years ago

Which formula correctly expresses the property density?

Physics

1 answer:

omeli [17]3 years ago

6 0

Answer:

Explanation:

density is defined to be as the mass per unit volume therefore you divid the mass by the volume

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A rifle fires a 3.56 x 10-2-kg pellet straight upward, because the pellet rests on a compressed spring that is released when the

vivado [14]

Answer:

k = 773.64 N/m

Explanation:

Given:

mass of the pellet, m = 3.56 x 10² kg

Compression in spring, x = 6.61 × 10⁻² m

Height of rise of the pellet, h = 4.83 m

Now, let k be the spring constant

From the concept of conservation of energy, we get

gravitation potential energy acquired by the pellet will be from the energy provided by the spring

thus,

$mgh = \frac{1}{2}kx^2$

where, g is the acceleration due to the gravity

on substituting the values, we get

$3.56\times10^2\times9.8\times4.83 = \frac{1}{2}k\times(6.61\times10^{-2})^2$

1.685 = .002178 × k

k = 773.64 N/m

5 0

3 years ago

Talking to adults can be intimidating and uncomfortable if you do not yet feel like one. However, learning how to effectivelycom

Korvikt [17]

Answer: i didnt look at your question its to long

Explanation:

8 0

4 years ago

A car collides head on with a stationary sign. The collision brings the car to a stop, but the sign does not move.

olga_2 [115]

A.a sign that breaks loose from the ground when a force is applied to it

7 0

3 years ago

What must be the distance (in meters) between a point charge q1 = 16 μC and a point charge q2 = 32 μC for the electrostatic forc

adoni [48]

Answer:

d = 0.71 meters

Explanation:

It is given that,

Charge 1, $q_1=16\ \mu C=16\times 10^{-6}\ C$

Charge 2, $q_2=32\ \mu C=32\times 10^{-6}\ C$

Electrostatic force between charges, F = 9 N

Let d is the distance between the charges. The electrostatic force between the charges is given by the product of charges and divided by square of distance between them. Mathematically, it is given by :

$F=k\dfrac{q_1q_2}{d^2}$

$d=\sqrt{\dfrac{kq_1q_2}{F}}$

$d=\sqrt{\dfrac{9\times 10^9\times 16\times 10^{-6}\times 32\times 10^{-6}}{9}}$

d = 0.71 meters

So, the distance between the charges is 0.71 meters. Hence, this is the required solution.

6 0

4 years ago

You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics

spayn [35]

Answer:

$v_0 = 3.53~{\rm m/s}$

Explanation:

This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.

The initial velocity is in the x-direction, and there is no acceleration in the x-direction.

On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.

Applying the equations of kinematics in the x-direction gives

$x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}$

For the y-direction gives

$v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t$

Combining both equation yields the y_component of the final velocity

$v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}$

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

$\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}$

6 0

3 years ago