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3 years ago

8

An object is moving at a velocity of 23 m/s. It accelerates to a velocity of 85 m/s over a time of 2.0 s. What acceleration did

it cynorionan?

1 answer:

maw [93]3 years ago

3 0

I believe its 31m/s^2

Explanation:

vf-vi/t

85-23/2

85-23=62

62/2=31

31m/s^2

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2. A ball is thrown 45 m into the air. It takes the ball 5 s to hit the ground. What is the

koban [17]

Answer:

9m/s

Explanation:

45m÷5s=9m/s

Hope it helps!!!

6 0

3 years ago

Read 2 more answers

) Force F = − + ( 8.00 N i 6.00 N j ) ( ) acts on a particle with position vector r = + (3.00 m i 4.00 m j ) ( ) . What are (a)

natali 33 [55]

To develop this problem it is necessary to apply the concepts related to the Cross Product of two vectors as well as to obtain the angle through the magnitude of the angles.

The vector product between the Force and the radius allows us to obtain the torque, in this way,

$\tau = \vec{F} \times \vec{r}$

$\tau = (8i+6j)\times(-3i+4j)$

$\tau = (8*4)(i\times j)+(6*-3)(j\times i)$

$\tau = 32k +18k$

$\tau = 50 k$

Therefore the torque on the particle about the origen is 50k

PART B) To find the angle between two vectors we apply the definition of the dot product based on the vector quantities, that is,

$cos\theta = \frac{r\cdot F}{|\vec{r}|*|\vec{F}|}$

$cos\theta = \frac{(8*-3)+(4*3)}{\sqrt{(-3)^2+4^2}*\sqrt{8^2+6^2}}$

$cos\theta = -0.24$

$\theta = cos^{-1} (-0.24)$

$\theta = 103.88\°$

Therefore the angle between the ratio and the force is 103.88°

5 0

3 years ago

1. A proton has a ____ charge <br> Positive <br> Negative <br> Or neutral

Westkost [7]

Answer:

Positive

Explanation:

Protons have positive charges

Electrons have negative charges and

Neutrons have neutral charges.

6 0

3 years ago

Read 2 more answers

Which of the following is the equivalent resistance of the circuit shown?

erica [24]

I believe the answer is going to B I hope this helps and good luck

6 0

3 years ago

Calculate the electric force an electron exerts upon a proton inside a He atom if they are d=2.7⋅10^-10m apart.

maksim [4K]

Explanation:

Charge of electron in He, $q_e=1.6\times 10^{-19}\ kg$

Charge of proton in He, $q_p=1.6\times 10^{-19}\ kg$

Distance between them, $d=2.7\times 10^{-10}\ m$

We need to find the electric force between them. It is given by :

$F=k\dfrac{q_eq_p}{d^2}$

$F=-9\times 10^9\times \dfrac{(1.6\times 10^{-19}\ C)^2}{(2.7\times 10^{-10}\ m)^2}$

$F=-3.16\times 10^{-9}\ N$

Since, there are two protons so, the force become double i.e.

$F=2\times 3.16\times 10^{-9}\ N$

$F=6.32\times 10^{-9}\ N$

So, the correct option is (c). Hence, this is the required solution.

6 0

3 years ago