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3 years ago

A football is thrown with an acceleration of 15 m/s^2 and a force of 13 N. What is its mass?

Physics

2 answers:

-Dominant- [34]3 years ago

7 0

I believe it is b. Lmk if I’m wrong

nika2105 [10]3 years ago

7 0

B. 0.87 kg because 13 N and m/s^2

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What is the correct water depth for an echo travel time of 6 seconds? (for the purposes of this exercise, assume pressure and te

saul85 [17]

In physics, there are empirical values for common important parameters. For example, the speed of sound is equal to 340 meters per second. Unlike the speed of light, the speed of sound is dependent on temperature and pressure. But in standard room conditions, the speed is 340 m/s. Using this value of speed, we can find the depth given the time. You should also note that the distance the echo travels is exactly the same distance that object travelled. Therefore,

Distance = 340 m/s * 6 seconds
Distance = 2,040 meters deep

7 0

4 years ago

The conventional relatively small unit for work(ignoring time) such as raising one pound one foot is the foot-pound(ft. lb.). Si

Lelechka [254]

Answer:

1 joule = 0.737 foot-pound

Joule is the unit of work.

1 J = 1 N·m

In SI units

1 J = 1 kg· m/s²

0.737 foot-pound is the amount of work to raise 0.737 pounds one foot or raising one pound to 0.737 ft.

6 0

3 years ago

The Pacific Plate is moving 29 mm/year toward the north and 20 mm/year toward the west relative to the North American Plate. Sho

adell [148]

Answer:

7 mm per year

Explanation:

It is given that :

The Pacific plate is moving towards north at = 29 mm per year

The Pacific plate is moving towards west at = 20 mm per year

We have to calculate the total relative motion towards the northwest.

So we have to find the resultant of the two motions.

Since the two movements are perpendicular, therefore the angle between the two motions is 90 degree.

Therefore, finding their resultant,

$R^2=(29)^2+(20)^2$

$R^2=(49)^2$

R = 7

Therefore, total relative motion towards the northwest is 7 mm per year.

4 0

3 years ago

You take a Frisbee to the top of the Washington Monument and send it sailing along horizontally at a speed of 5 m/s from the ver

Dominik [7]

This question can be solved with the help of the equations of motion.

A) The Frisbee will remain in the air for "5.87 s".

B) The frisbee will go "29.4 m" down the range.

To calculate the time, the frisbee will remain in the air, we will use the second equation of motion, for the vertical motion.

$h = v_it+\frac{1}{2}gt^2\\\\$

where,

h = height = 169.2 m

vi = initial velocity's vertical component = 0 m/s

g = acceleration due to gravity = 9.81 m/s²

t = time = ?

Therefore,

$169.2\ m = (0\ m/s)(t)+\frac{1}{2}(9.81\ m/s^2)t^2\\\\t = \sqrt{\frac{(169.2\ m)(2)}{9.81\ m/s^2}}$

t = 5.87 s

Now, we will calculate the horizontal range by applying the equation for constant motion. Because the velocity in the horizontal direction will remain constant due to no air resistance

s = vt

where,

s = horizontal range = ?

v= initial velocity's horizontal component = 5 m/s

t = time = 5.87 s

Therefore,

s = (5 m/s)(5.87 s)

s = 29.4 m

Learn more about equations of motion here:

brainly.com/question/9772550?referrer=searchResults

8 0

3 years ago

a 2.6 kg block is attached to a horizontal spring that has a spring constant of 126 N/m. At the instant when the displacement of

NeTakaya

Answer:

The acceleration of the object is $5.57\ m/s^2$ .

Explanation:

Given that,

Mass of the block, m = 2.6 kg

Spring constant of the spring, k = 126 N/m

At the instant when the displacement of the spring from its unstained length is 0.115 m. We need to find the acceleration of the object.

When the block is displaced, the force acting on the spring is equal to the force due to its motion. Such as :

$kx=ma$

a is acceleration of the object

$a=\dfrac{kx}{m}\\\\a=\dfrac{126\times 0.115}{2.6}\\\\a=5.57\ m/s^2$

So, the acceleration of the object is $5.57\ m/s^2$ .

8 0

3 years ago