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3 years ago

Whwhich pf the fallowing equations can be used to directly calculate an objects impulse

Physics

2 answers:

Bad White [126]3 years ago

8 0

Since I don't see a picture, the impulse is
p=m*v
where m is the mass
and is the velocity

sashaice [31]3 years ago

5 0

P = mass times velocity need to show a pic to get help

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A certain bridge is 4,224 feet long. What constant rate, in miles per hour, must be maintained in order to walk across the bridg

Archy [21]

Answer:

4miles/hour

Explanation:

the solution for this question requires that the quantities are converted to the appropriate units as required by the question.

Rate in miles per hour = distance in miles / time in hour

to convert 12 minutes to hours; recall that 60 minutes make 1 hour

12 minutes to hour = 12/60 = 0.2hr

to convert 4224 feet to miles; recall 5280 feet is equivalent to 1 mile

4224 feet to miles = 4224/5280 = 0.8 miles

∴ rate = 0.8 / 0.2

rate = 4 miles per hour

the constant rate in miles per hour = 4 miles/hour

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3 years ago

What type of simple machine is an electric fan?

GrogVix [38]

Answer:

An electric fan is considered to be a mixture of several simple machines. It includes the Wheel and Axle type, wedges, and the Inclined plane types. The blades of an electric fan are the inclined planes and the wedges.

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2 years ago

What do you think would happen to the force of attraction of two interacting charges if their distance apart is halved?

sweet [91]

Answer:

The new force becomes 4 times the initial force.

Explanation:

The force of attraction or repulsion is given by the relation as follows :

$F=k\dfrac{q_1q_2}{d^2}$

Where

d is the distance between the interacting charges

F is inversely proportional to the distance between charges.

If the distance is halved, d'=(d/2), new force is given by :

$F'=k\dfrac{q_1q_2}{d'^2}\\\\=k\dfrac{q_1q_2}{(\dfrac{d}{2})^2}\\\\=k\dfrac{q_1q_2}{\dfrac{d^2}{4}}\\\\=4\times \dfrac{kq_1q_2}{d^2}\\\\F'=4F$

So, the new force becomes 4 times the initial force.

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2 years ago

You toss a ball straight up with an initial speed of 30m/s. How high does it go, and how long is it in the air (neglecting air r

Brut [27]

Explanation:

Given that,

A ball is tossed straight up with an initial speed of 30 m/s

We need to find the height it will go and the time it takes in the air.

At its maximum height, its final speed, v = 0 and it will move under the action of gravity. Using equation of motion :

v = u +at

Here, a = -g

v = u -gt

i.e. u = gt

$t=\dfrac{u}{g}\\\\t=\dfrac{30\ m/s}{9.8\ m/s^2}\\\\t=3.06\ s$

So, the time for upward motion is 3.06 seconds. It means that it will in air for 3.06×2 = 6.12 seconds

Let d is the maximum distance covered by it.

$d=ut-\dfrac{1}{2}gt^2$

Putting all values

$d=30(3.06)-\dfrac{1}{2}\times 9.8\times (3.06)^2\\\\d=45.91\ m$

Hence, it will go to a height of 45.91 m and it will in the air for 6.12 seconds.

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2 years ago

A spring does 80 J of work launching a 1.85 kg rock into the air. Ignoring air resistance, how high will the rock go?

Svetlanka [38]

h=80/(1.85*9.8)=4.4 m

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2 years ago