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3 years ago

What is the difference between mass and weight?

Physics

1 answer:

xxTIMURxx [149]3 years ago

4 0

Answer:

C. Mass is independent of the force acting on the mass.

Weight depends on the gravitational force acting on the mass.

Example: A mass of 1 kg on the earth still has a mass of 1 kg on the moon;

whereas the weight of that mass on earth is 9.8 N and only 1/6 of that amount on the moon.

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Suppose that you can throw a projectile at a large enough v0 so that it can hit a target a distance R downrange. Given that you

NikAS [45]

Answer:

Theta1 = 12° and theta2 = 168°

The solution procedure can be found in the attachment below.

Explanation:

The Range is the horizontal distance traveled by a projectile. This diatance is given mathematically by Vo cos(theta) t. Where t is the total time of flight of the projectile in air. It is the time taken for the projectile to go from starting point to finish point. This solution assumes the projectile finishes uts motion on the same horizontal level as the starting point and as a result the vertical displacement is zero (no change in height).

In the solution as can be found below, the expression to calculate the range for any launch angle theta was first derived and then the required angles calculated from the equation by substituting the values of the the given quantities.

7 0

3 years ago

A garden hose has a radius of 0.0120 m, and water initially comes out at a speed of 2.88m/s. Dasha puts her thumb over the end ,

creativ13 [48]

Answer:

v = 12.4 [m/s]

Explanation:

With the speed and Area information, we can determine the volumetric flow.

$V=v*A\\A=\pi *r^{2}$

where:

r = radius = 0.0120 [m]

v = 2.88 [m/s]

$A=\pi *(0.0120)^{2} \\A=4.523*10^{-4} [m]\\$

Therefore the flow is:

$V=2.88*4.523*10^{-4} \\V=1.302*10^{-3} [m^{3}/s ]$

Despite the fact that you cover the inlet with the finger, the volumetric flow rate is the same.

$v=V/A\\v=1.302*10^{-3} /1.05*10^{-4} \\v=12.4[m/s]$

3 0

3 years ago

A ball is dropped from a building of height h. Assume the ball starts from rest and that air friction can be ignored. Derive an

agasfer [191]

Answer:

$t=\sqrt{h/g}$

Explanation:

We use the kinematics equation to solve this question:

$y(t)=y_{o}+v_{o}t+1/2*a*t^{2}$

$v_{o}=0$ because the ball is dropped

$a=-g$ the acceleration is the gravity, negative because it points downwards

$y_{o}=h$ initial height

$y(t)=h/2$ final height

So:

$h/2=h-1/2*g*t^{2}$

$t=\sqrt{h/g}$

8 0

3 years ago

An astronaut lands on a new, recently discovered planet in a different star system. The astronaut measures the acceleration due

Nezavi [6.7K]

Answer:

The radius of the new planet is ~2.04 * 10⁶ m, or 2,041,752 m.

Explanation:

We can use Newton's Law of Universal Gravitation:

$\displaystyle F_g=G\frac{Mm}{r^2}$

Let's look at Newton's 2nd Law:

$F=ma$

We can set these equations equal to each other:

$\displaystyle G\frac{Mm}{r^2} =ma$

The mass of the second mass (astronaut) cancels out. We are left with:

$\displaystyle G\frac{M}{r^2} =a$

We are solving for the radius of the new planet, so we can rearrange the equation:

$\displaystyle r=\sqrt{\frac{GM}{a} }$

Substitute in our known values given in the problem (G = 6.67 * 10⁻¹¹ ; M = 7.5 * 10²³ ; a = 12).

$\displaystyle r =\sqrt{\frac{(6.67\times 10^{-11})(7.5 \times 10^{23}}{12} }$
$r=2.04 \times 10^6$

The radius of the new planet is ~2.04 * 10⁶ m.

3 0

2 years ago

A. What is an electric field line?

Lorico [155]

Answer:

Please see below as the answers are self-explanatory

Explanation:

A electric field line is an imaginary line, which has the property that the electric field vector is tangent to it at any point. It starts from positive charges (since the electric field by convention it has the direction of the trajectory that would take a positive test charge, so it always goes away from positive charges) and ends in negative charges.

Since the potential difference between two points represents the work per unit charge needed for a charge to move between these points, a equipotential surface is the one over which it is not needed to do work to move a charge from any point on the surface to any other point, which means that all points are at the same potential.

Equipotential surfaces are not necessarily physical surfaces, they can be defined in vaccum for instance.
As an example, any spherical surface concentric with a point charge, is an equipotential surface, and it can be a real surface or a fictitious one.

7 0

3 years ago