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3 years ago

6

_ is amount of “ground” an object moves from its starting point and __ is how much “ground” and object covers in

total.
Question options:

Displacement, distance

Speed, displacement

Distance, displacement

Movement, distance

2 answers:

Kamila [148]3 years ago

8 0

The correct answer is C.

Papessa [141]3 years ago

5 0

I believe the question talks about displacement and distance. The difference of the two is that displacement is the linear distance or shortest distance drawn from the starting point until the final point while distance is the whole trip that is covered, so the answer is:

<span>Displacement, distance</span>

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What is the electric potential energy of an electron at the negative end of the cable, relative to the positive end of the cable

VashaNatasha [74]

Answer:

Electric potential energy at the negative terminal: $1.92\cdot 10^{-18}J$

Explanation:

When a particle with charge $q$ travels across a potential difference $\Delta V$ , then its change in electric potential energy is

$\Delta U = q \Delta V$

In this problem, we know that:

The particle is an electron, so its charge is

$q=-1.60\cdot 10^{-19}C$

We also know that the positive terminal is at potential

$V_+=0V$

While the negative terminal is at potential

$V_-=-12 V$

Therefore, the potential difference (final minus initial) is

$\Delta V = -12-0 = -12 V$

So, the change in potential energy of the electron is

$\Delta U = (-1.6\cdot 10^{-19})(-12)=1.92\cdot 10^{-18}J$

This means that the electron when it is at the negative terminal has $1.92\cdot 10^{-18}J$ of energy more than when it is at the positive terminal.

Since the potential at the positive terminal is 0, this means that the electric potential energy of the electron at the negative end is

$1.92\cdot 10^{-18}J$

3 0

3 years ago

A person kicks a 4.0-kilogram door with a 48-newton force causing the door to accelerate at 12 meters per second squared. What i

inna [77]

Answer:

-48 N

Explanation:

mass of door (m) = 4 kg

acceleration of the door = 12 m/s^{2}

force exerted by the person = 48 N

From Newton's third law of motion, action and reaction are equal but opposite. Therefore the force exerted on the door by the person which is 48 N will be the same as the force exerted on the person by the door but opposite in its direction, and this would be - 48 N

7 0

2 years ago

At its widest point, the diameter of a bottlenose dolphin is 0.50 m. Bottlenose dolphins are particularly sleek, having a drag c

fiasKO [112]

Answer:

497.00977 N

3742514.97005

Explanation:

$\rho$ = Density of water = 1000 kg/m³

C = Drag coefficient = 0.09

v = Velocity of dolphin = 7.5 m/s

r = Radius of bottlenose dolphin = 0.5/2 = 0.25 m

A = Area

Drag force

$F_d=\frac{1}{2}\rho CAv^2\\\Rightarrow F_d=\frac{1}{2}\times 1000 \times 0.09(\pi 0.25^2)7.5^2\\\Rightarrow F_d=497.00977\ N$

The drag force on the dolphin's nose is 497.00977 N

at 20°C

$\mu$ = Dynamic viscosity = $1.002\times 10^{-3}\ Pas$

Reynold's Number

$Re=\frac{\rho vd}{\mu}\\\Rightarrow Re=\frac{1000\times 7.5\times 0.5}{1.002\times 10^{-3}}\\\Rightarrow Re=3742514.97005$

The Reynolds number is 3742514.97005

8 0

3 years ago

Two asteroids identical to those above collide at right angles and stick together; i.e, their initial velocities were perpendicu

11111nata11111 [884]

Answer:

velocity = 62.89 m/s in 58 degree measured from the x-axis

Explanation:

Relevant information:

Before the collision, asteroid A of mass 1,000 kg moved at 100 m/s, and asteroid B of mass 2,000 kg moved at 80 m/s.

Two asteroids moving with velocities collide at right angles and stick together. Asteroid A initially moving to right direction and asteroid B initially move in the upward direction.

Before collision Momentum of A = 1000 x 100 = $10^5$ kg - m/s in the right direction.

Before collision Momentum of B = 2000 x 80 = 1.6 x $10^5$ kg - m/s in upward direction.

Mass of System of after collision = 1000 + 2000 = 3000 kg

Now applying the Momentum Conservation, we get

Initial momentum in right direction = final momentum in right direction = $10^5$

And, Initial momentum in upward direction = Final momentum in upward direction = 1.6 x $10^5$

So, $V_x = \frac{10^5}{3000}$ = $\frac{100}{3}$ m/s

and $V_y=\frac{160}{3}$ m/s

Therefore, velocity is = $\sqrt{V_x^2 + V_y^2}$

= $\sqrt{(\frac{100}{3})^2 + (\frac{160}{3})^2}$

= 62.89 m/s

And direction is

tan θ = $\frac{V_y}{V_x}$ = 1.6

therefore, $\theta = \tan^{-1}1.6$

= $58 ^{\circ}$ from x-axis

4 0

3 years ago

When landing from a jump, a basketball player of mass 82 kg has a velocity of 1.2 m/s right before they hit the ground. The play

saw5 [17]

Answer:

2361.6N

Explanation:

Mass of player = 82kg

Velocity = 1.2m/s

Kinetic energy of player:

= 1/2mv²

= 1/2*82*1.2²

= 41x1.44

= 59.04J

Final kinetic energy = 0

Change in kinetic energy

|∆k| = |0-59.04|

= 59.04

Workdone by the feet = fd

d = 0.025

Fd = 59.04

F = 59.04/0.025

= 2361.6N

This is his average force.

6 0

2 years ago