All categories

3 years ago

Use the motion map to answer the question.

Physics

1 answer:

Lana71 [14]3 years ago

3 0

Answer:

The object starts away from the origin and then moves toward the origin at a constant velocity. Next, it stops for one second. Finally, it moves away from the origin at a greater constant velocity.

You might be interested in

Is electrical conductivity?

expeople1 [14]

Answer:

can't tell if this is question, it is not written correctly

Explanation:

Electrical conductivity is the measure of a material's ability to allow the transport of an electric charge. Its SI is the siemens per meter, (A2s3m−3kg−1) (named after Werner von Siemens) or, more simply, Sm−1. It is the ratio of the current density to the electric field strength.

8 0

3 years ago

Which of the following is an example of a force<br> inertia<br> push <br> pull

Whitepunk [10]

A force can be considered a push or pull

hope this helps :)

8 0

3 years ago

300,000,000m/s × 2 = —— m

nekit [7.7K]

Answer:

6 x 10⁸ m

Explanation:

300,000,000 m/s x 2 s

= 3 x 10⁸ x 2

= 6 x 10⁸ m

7 0

3 years ago

A 79.7 kg base runner begins his slide into second base while moving at a speed of 4.77 m/s. The coefficient of friction between

SashulF [63]

To solve this problem we will apply the concept related to the kinetic energy theorem. Said theorem states that the work done by the net force (sum of all forces) applied to a particle is equal to the change experienced by the kinetic energy of that particle. This is:

$\Delta W = \Delta KE$

$\Delta W = \frac{1}{2} mv^2$

Here,

m = mass

v = Velocity

Our values are given as,

$m = 79.7kg$

$v = 4.77m/s$

Replacing,

$\Delta W = \frac{1}{2} (79.7kg)(4.77m/s)^2$

$\Delta W = 907J$

Therefore the mechanical energy lost due to friction acting on the runner is 907J

6 0

3 years ago

A proton, traveling with a velocity of 3.7 × 106 m/s due east, experiences a magnetic force that has a maximum magnitude of 5.4

Vikki [24]

The magnetic force (Lorentz force) experienced by the proton in the magnetic field is given by
$F=qvBsin\theta=qvB$
since $\theta = 90^{\circ}$ , because the velocity v and the force F in this problem are perpendicular, and so also the angle $\theta$ between the velocity and the magnetic field B should be $90^{\circ}$ .

Let's find the magnitude of the magnetic field; this is given by
$B= \frac{F}{qv}= \frac{5.4\cdot 10^{-14}N}{1.6\cdot 10^{-19}C \cdot 3.7\cdot 10^6 m/s}=0.091 T$

To understand the direction, let's use the right-hand rule:
-index finger: velocity
- middle finger: magnetic field
- thumb: force

Since the velocity (index) points east and the force (thumb) points south, then the magnetic field (middle finger) points downwards. So we write:
B = -0.091 T

8 0

3 years ago