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Which phrase describes speed?

Anna11 [10]

Answer:

the answer is B

Explanation:

speed is the rate at which the distance covered changes or the distance divided by the time taken.

scalar is always positive.

6 0

2 years ago

Read 2 more answers

A circular curve of radius 150 m is banked at an angle of 15 degrees. A 750-kg car negotiates the curve at 85.0 km/h without ski

Crazy boy [7]

Answer: a) 7.1 * 10^3 N; b) -880 N directed out of the curve.

Explanation: In order to solve this problem we have to use the Newton laws, then we have the following:

Pcos 15°-N=0

Psin15°-f= m*ac

from the first we obtain N, the normal force

N=750Kg*9.8* cos (15°)= 7.1 *10^3 N

Then to calculate the frictional force (f) we can use the second equation

f=P sin (15°) -m*ac where ac is the centripetal acceletarion which is equal to v^2/r

f= 750 *9.8 sin(15°)-750*(85*1000/3600)^2/150= -880 N

6 0

3 years ago

The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m

Simora [160]

Answer:

The distance between the charges is 5,335.026 m

Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

$F = k \frac{q_1 q_2}{d^2}$

where k is Coulomb's constant, $q_1$ and $q_2$ are the charges and d is the distance between the charges.

Working a little the equation, we can take:

$d^2 = k \frac{q_1 q_2}{F}$

$d = \sqrt{ k \frac{q_1 q_2}{F}}$

And this equation will give us the distance between the charges. Taking the values of the problem

$k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00$

(the force has a minus sign, as its attractive)

$d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}$

$d = \sqrt{ 28,462,500 \ m^2}}$

$d = 5,335.026 m$

And this is the distance between the charges.

3 0

3 years ago

A small, 300 g cart is moving at 1.20 m/s on an air track when it collides with a larger, 2.00 kg cart at rest?

stiv31 [10]

Answer:

The speed of the large cart after collision is 0.301 m/s.

Explanation:

Given that,

Mass of the cart, $m_1 = 300\ g = 0.3\ kg$

Initial speed of the cart, $u_1=1.2\ m/s$

Mass of the larger cart, $m_2 = 2\ kg$

Initial speed of the larger cart, $u_2=0$

After the collision,

Final speed of the smaller cart, $v_1=-0.81\ m/s$ (as its recolis)

To find,

The speed of the large cart after collision.

Solution,

Let $v_2$ is the speed of the large cart after collision. It can be calculated using conservation of momentum as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$m_1u_1+m_2u_2-m_1v_1=m_2v_2$

$v_2=\dfrac{m_1u_1+m_2u_2-m_1v_1}{m_2}$

$v_2=\dfrac{0.3\times 1.2+0-0.3\times (-0.81)}{2}$

$v_2=0.301\ m/s$

So, the speed of the large cart after collision is 0.301 m/s.

4 0

3 years ago

Write a list of at least five scientific methods.

kolbaska11 [484]

1- You should always have a question for your experiment.
2- You need to conduct background research. It helps to write down your sources so you can cite your references.
3- Propose a hypothesis (educated guess on what you believe the outcome of the experiment will be)
4- Design and perform an experiment to test your hypothesis (include independent and dependent variable)
5- Record observations and analyze what the data means.
6- Conclude whether you need to accept or reject your hypothesis, which accepting means your hypothesis was right and rejected is if it was wrong.

3 0

3 years ago