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yKpoI14uk [10]
4 years ago
7

How much force is needed to accelerate a 1750-kg car at a rate of 3 m/s2?

Physics
1 answer:
Rina8888 [55]4 years ago
5 0
The force needed is 5250N
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A student designs an electromagnet, as shown in the picture. The electromagnet is only able to pick up 1 paper clip. List 2 modi
Lerok [7]

-- Change the battery to one with higher voltage.

-- Wrap more turns of wire around the spike.

-- If the spike is made of anything else but iron, replace it with a pure iron one.

6 0
3 years ago
An organism’s scientific name consists of a. its class name and its family name. b. its kingdom name and its phylum name. c. its
mr Goodwill [35]
An organism scientific name consist of : C. its genus name and its species name
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hope this helps

7 0
3 years ago
A ball droped from a building. How fast is it traveling after falling 3.55s
algol [13]

Answer:

d = 61.75 m

Explanation:

Given that,

A ball droped from a building.

We need to find how fast is it traveling after falling 3.55 s.

As it is dropped, its initial velocity is equal to 0.

Let d is the distance it covers after falling 3.55 s.

We can use second equation of motion to find d.

d=ut+\dfrac{1}{2}at^2

Here, u = 0 and a =g

d=\dfrac{1}{2}gt^2\\\\d=\dfrac{1}{2}\times 9.8\times (3.55)^2\\\\d=61.75\ m

So, it will cover 61.75 m after falling 3.55 seconds.

4 0
3 years ago
Was the cannonball able to hit its target of 50 meters when the initial velocity was 20 m/s? Why?/Why Not?
frosja888 [35]

Answer:

The cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.

Explanation:

From the question given above, the following data were obtained:

Height to which the target is located = 50 m

Initial velocity (u) = 20 m/s

To know whether or not the cannon ball is able to hit the target, we shall determine the maximum height to which the cannon ball attained. This can be obtained as follow:

Initial velocity (u) = 20 m/s

Final velocity (v) = 0 (at maximum height)

Acceleration due to gravity (g) = 10 m/s²

Maximum height (h) =?

v² = u² – 2gh (since the ball is going against gravity)

0² = 20² – (2 × 10 × h)

0 = 400 – 20h

Collect like terms

0 – 400 = – 20h

– 400 = – 20h

Divide both side by – 20

h = – 400 / – 20

h = 20 m

Thus, the the maximum height to which the cannon ball attained is 20 m.

From the calculations made above, we can conclude that the cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.

3 0
3 years ago
Consider the reaction, X + 2Y → XY2 If X and Y are completely consumed in the reaction and we start with 10.0 mol of Y, then how
Harrizon [31]

Answer:

In the reaction you would have 15.0 mols of Y and X.

Explanation:

The stoichiometric coefficents for X and Y are 1 and 2 respectively, if you start the reaction with 10.0 moles of Y you would need 5.0 moles of X in order to achieve a complete reaction so you will have 15.0 total moles in the reaction, assuming no mass loss and no nuclear reactions.

3 0
3 years ago
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