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2 years ago

Find average speed when time= 27s and total distance is 94m

Physics

1 answer:

Slav-nsk [51]2 years ago

8 0

3.4814815 (or 3 13/27) m/s

speed = distance/time
3.4814815 (or 3 13/27) = 94/27

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Which of these statements is true about endothermic reactions, but not about exothermic reactions?

sashaice [31]

I think the correct answer from the choices listed above is the second option. For endothermic reactions, the reactants have less energy than the products. Which would mean that energy should be added to the reaction for it to proceed. Hope this answers the question.

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2 years ago

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Name the net force between objects that allows a car to travel in a circle

iren [92.7K]

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3 years ago

Vector ~A has a magnitude of 29 units and points in the positive y-direction. When vector ~B is added to ~A, the resultant vecto

timurjin [86]

Answer:

The magnitude of vector B is 43 units and it points in the negative y-direction.

Explanation:

Resultant of vectors = vector sum of all the vectors

Vector A = 29j

Vector B = ?

Resultant of vector A and B = R = -14j

R = A + B

-14j = 29j + B

B = -14j - 29j = - 43j

Hence, the magnitude of vector B is 43 units and it points in the negative y-direction.

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2 years ago

A load of mass 5kg is raised through a height of 2m. calculate the work done against (g=10mls)

Illusion [34]

The work done against gravity is 100 J

Explanation:

The work done against gravity in order to lift an object is equal to the change in gravitational potential energy of the object:

$W=mg\Delta h$

where

m is the mass of the object

g is the acceleration of gravity

$\Delta h$ is the change in height of the object

For the object in this problem, we have:

m = 5 kg

$g=10 m/s^2$

$\Delta h = 2 m$

Substituting into the equation,

$W=(5)(10)(2)=100 J$

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

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3 years ago

Each of the following diagrams shows a spaceship somewhere along the way between Earth and the Moon (not to scale); the midpoint

djyliett [7]

Answer:

$F_5 >F_4>F_1 >F_2>F_3$

Where $F_i$ represent the force for each of the 5 cases $i -1,2,3,4,5$ presented on the figure attached.

Explanation:

For this case the figure attached shows the illustration for the problem

We have an inverse square law with distance for the force, so then the force of gravity between Earth and the spaceship is lower when the spaceship is far away from Earth.

Th formula is given by:

$F = G \frac{m_{Earth} m_{Spaceship}}{r^2}$

Where G is a constant $G = 6.674 x10^{-11} m^2/ (ks s^2)$

$m_{Earth}$ represent the mass for the earth

$m_{spaceship}$ represent the mass for the spaceship

$r$ represent the radius between the earth and the spaceship

For this reason when the distance between the Earth and the Spaceship increases the Force of gravity needs to decrease since are inversely proportional the force and the radius, and for the other case when the Earth and the spaceship are near then the radius decrease and the Force increase.

Based on this case we can create the following rank:

$F_5 >F_4>F_1 >F_2>F_3$

Where $F_i$ represent the force for each of the 5 cases $i -1,2,3,4,5$ presented on the figure attached.

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2 years ago