All categories

3 years ago

Tyler and his family drove from Atlanta to Baltimore. They traveled a total of 678 miles over 2 days.

Physics

2 answers:

sweet-ann [11.9K]3 years ago

7 0

Answer:

Atlanta

Explanation:

Keith_Richards [23]3 years ago

6 0

Answer:

atlanta

Explanation:

just took the test

You might be interested in

How do you make magnetic and static forces strong or week?

victus00 [196]

You make them strong for example if you rub a balloon in your hair the longer you do it the longer it sticks because of static

6 0

3 years ago

Which statement is true according to Hooke's law?

Mkey [24]

Hooke's law:

F = -kx

F is the spring force, k is the spring constant, and x is the displacement of the spring. The negative sign is there to indicate that the force acts in the opposite direction of the displacement (e.g. if you stretch a spring towards the right, you will feel a force pulling towards the left).

Looking at Hooke's law, the magnitude of the force is directly proportional to its displacement; that is to say, if the displacement increases, so does the magnitude of the spring force, and if the magnitude of the force increases, this necessitates an increase in the displacement.

1. As the force on a spring increases, the distance it stretches decreases.

An increase in force necessitates an increase in displacement. This statement is false.

2. As the force on a spring increases, the distance it stretches increases.

An increase in force necessitates an increase in displacement. This statement is true.

3. As the force on a spring increases, the spring's length remains unchanged.

An increase in force necessitates an increase in displacement. This statement is false.

4. The distance a spring stretches has nothing to do with how much force is applied to it.

Hooke's law shows that the spring force and its displacement are very much dependent on each other. This statement is false.

Only statement 2 is true. As the force on a spring increases, the distance it stretches increases.

3 0

3 years ago

T/-15=-3. what is t?

sweet-ann [11.9K]

Answer:

$\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}$

4 0

2 years ago

Supposing d(t) is known to have value D,

creativ13 [48]

Answer:

The procedure is: solve the quadratic equation for $t$ .

Explanation:

This question assumes uniformly accelerated motion, for which the distance d a particle travels in time t is given by the general equation:

$d(t)=d_0+v_0t+at^2/2$

That is a quadratic equation, where the independent variable is the time $t$ .

Thus, the procedure that will find the time t at which the distance value is known to be D is to solve the quadratic equation for $t$ .

To solve it you start by changing the equation to the general form of the quadratic equations, rearranging the terms:

$(a/2)t^2+v_0t+(d_0-D)=0$

Some times that equation may be solved by factoring, and always it can be solved by using the quadratic formula:

$t=\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

Where:

$a=-a/2\\ \\ b=v_0\\ \\ c=d_0-D$

That may have two solutions. Some times one of the solution makes no physical sense (for example time cannot be negative) but others the two solutions are valid.

5 0

3 years ago

The distance traveled by the scooter driver from 9:00 to 11:00 am is

wlad13 [49]

B is the correct answer

6 0

3 years ago