New ocean crust or sea floor is produced when tectonic plates ______

What is the acceleration of a 10 kg mass pushed 5 n forceushed by a 5n(kg-m/s2) (use gresa method)

Pepsi [2]

In this problem,

Applied force(F) = 10 N

The object’s mass (m) is 5 kg.

Having said that,

An object’s force is equal to the product of its mass and the acceleration it experiences as a result of the applied force.

i.e., Mass + Acceleration = Force (a)

F= m×a

Therefore,

A= F÷m

A= (10÷5) m/sec²

A= 2 m/sec²

Consequently, the object’s acceleration,

A=2 m/sec²

Concept of force and acceleration:

This states that the rate of velocity change of an object is directly proportional to the applied force and moves in the direction of the applied force.

It can be expressed mathematically as force (N) = mass (kg) x acceleration (m/s2). Therefore, an object with constant mass will accelerate in direct proportion to the applied force.

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6 0

1 year ago

a student balances a 1.5kg of broom by placing her finger 1.4m from the end of the broom handle. How far from the broom handle w

Delicious77 [7]

The broom handle that she have to balance if she hung a 400g mass from the end of the broom handle is 5.24m

This problem is centered on moment. Moment is the turning effect of a force about a point. It is expressed as:

Moment = Force× Distance

According to principle of moment, the sum of clockwise moment is equal to sum of anticlockwise moment at shown

M1d1 = M2d2

Given the following

M1 = 1.5kg

d1 = 1.4m

M2 = 400g = 0.4kg

d2 is required

Substitute

1.5(1.4) = 0.4d2

2.1 = 0.4d2

d2 = 2.1/0.4

d2 = 5.24m

Hence the broom handle that she have to if she hung a 400g mass from the end of the broom handle is 5.24m

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4 0

2 years ago

A train travels due north in a straight line with a constant speed of 100 m/s. Another train leaves a station 2,881 m away trave

damaskus [11]

Answer:

The trains will collide at a distance 1660 m from the station

Explanation:

Let the train traveling due north with a constant speed of 100 m/s be Train A.

Let the train traveling due south with a constant speed of 136 m/s be Train B.

From the question, Train B leaves a station 2,881 m away (that is 2,881 m away from Train A position).

Hence, the two trains would have traveled a total distance of 2,881 m by the time they collide.

∴ If train A has covered a distance $x$ m by the time of collision, then train B would have traveled $(2881 - x)$ m.

Also,

At the position where the trains will collide, the two trains must have traveled for equal time, t.

That is, At the point of collision,

$t_{A} = t_{B}$

$t_{A}$ is the time spent by train A

$t_{B}$ is the time spent by train B

From,

$Velocity = \frac{Distance }{Time }\\$

$Time = \frac{Distance}{Velocity}$

Since the time spent by the two trains is equal,

Then,

$\frac{Distance_{A} }{Velocity_{A} } = \frac{Distance_{B} }{Velocity_{B} }$

${Distance_{A} = x$ m

${Distance_{B} = 2881 - x$ m

${Velocity_{A} = 100$ m/s

${Velocity_{B} = 136$ m/s

Hence,

$\frac{x}{100} = \frac{2881 - x}{136}$

$136(x) = 100(2881 - x)\\136x = 288100 - 100x\\136x + 100x = 288100\\236x = 288100\\x = \frac{288100}{236} \\x = 1220.76m\\$

$x$ ≅ 1,221 m

This is the distance covered by train A by the time of collision.

Hence, Train B would have covered (2881 - 1221)m = 1660 m

Train B would have covered 1660 m by the time of collision

Since it is train B that leaves a station,

∴ The trains will collide at a distance 1660 m from the station.

7 0

3 years ago

Suppose the ski patrol lowers a rescue sled carrying an injured skier, with a combined mass of 97.5 kg, down a 60.0-degree slope

Kitty [74]

a. 1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b.21,835 J work, in joules, is done by the rope on the sled this distance.

c. 23,170 J the work, in joules done by the gravitational force on the sled d. The net work done on the sled, in joules is 43,670 J.

<h3>What is friction work?</h3>

The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement

a. How much work is done by friction as the sled moves 28m along the hill?

ans. We use the formula:

friction work = -µ.mg.dcosθ

= -0.100 * 97.5 kg * 9.8 m/s² * 28 m * cos 60

= -1337.3 J

-1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b. How much work is done by the rope on the sled in this distance?

We use the formula:

Rope work = -m.g.d(sinθ - µcosθ)

rope work = - 97.5 kg * 9.8 m/s² * 28 m (sin 60 – 0.100 * cos 60)

= 26,754 (0.816)

= 21,835 J

21,835 J work, in joules, is done by the rope on the sled this distance.

c. What is the work done by the gravitational force on the sled?

By using the formula:

Gravity work = mgdsinθ

= 97.5 kg * 9.8 m/s² * 28 m * sin 60

= 23,170 J

23,170 J the work, in joules done by the gravitational force on the sled .

D. What is the total work done?

By adding all the values

work done = -1337.3 + 21,835 + 23,170

= 43,670 J

The net work done on the sled, in joules is 43,670 J.

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4 0

1 year ago

Help i feel stressful in work i feel like giving up

Phantasy [73]

Answer:

Keep going you are doing great, if you need a break go take a short walk

8 0

2 years ago

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New ocean crust or sea floor is produced when tectonic plates _______. diverge converge