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Ilia_Sergeevich [38]

3 years ago

10

A force is acting on a 3-kg object during 5 seconds. If the object initially has a speed of 1.5m/s and after applying the force

its speed is 4 m/s

1 answer:

uranmaximum [27]3 years ago

3 0

Answer:

1.5 N

Explanation:

You've left us to guess what the question is. I will Assume it is what's the force?

Givens

m = 3 kg

vi = 1.5 m/s

vf = 4 m/s

t = 5 seconds

Formula

F = m * (vf - vi)/t

Solution

F = 3 * (4 - 1.5) / 5

F = 1.5 N

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3 years ago

A vehicle of known mass accelerates along a straight path. According to Newton's second law of motion, what caused this accelera

Naily [24]

The answer would be:

C. An unbalanced force has acted on the vehicle.

The presence of an unbalanced force will accelerate an object, the second law of motion dictates this (Although not explicitly). Lets knock out the rest of the choices.

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Hope you got your answer here, although you did not ask for an explanation, maybe this will help you figure some of the other questions you have on your own.

6 0

4 years ago

Read 2 more answers

A plane is flying horizontally with speed

Kay [80]

Answer:

2. ahead of the package.

Explanation:

Using y' - y = ut - 1/2gt², we find the time, t it takes the package to hit the ground. So, u = initial vertical velocity of package = 0 m/s, y = initial position of package = 3970 m, y' = final position of package = 0 m, g = acceleration due to gravity = 9.8 m/s².

Substituting the variables into the equation, we have

y' - y = ut - 1/2gt²

0 m - 3970 m = 0t - 1/2 × (9.8 m/s²)t²

-3970 m = -(4.9m/s²)t²

t² = -3970 m ÷ -4.9 m/s²

t² = 810.2 s²

t = √810.2 s²

t = 28.5 s

Using v = u' + at, we find the horizontal acceleration of the plane. Since the initial horizontal velocity of the package is that of the plane, u' = 162 m/s, v = 0 m/s since the package stops and t = 28.5 s when the package stops.

So, a = (v - u')/t

a = (0 m/s - 162 m/s)/28.5 s

a = -162 m/s/28.5 s

a = -5.68 m/s²

Using v² = u'² + 2as, we find the horizontal distance ,s where the package stops.

So, s = (v² - u'²)/2a

Substituting the values of the variables, we have

s = ((0 m/s)² - (162 m/s)²)/(2 × -5.68 m/s²)

= - 162 m²/s²/(-11.36 m/s²)

= 14.26 m

The horizontal distance d the plane moves after releasing the package is d = u't = 162 m/s × 28.5 s = 4617 m

Since d = 4617 m > s = 14.26 m, the plane would be ahead of the package when the package hits the ground.

6 0

3 years ago

How do i figure out the net force of an object? explain please :)

Savatey [412]

Draw a free body diagram to help you visualize. draw the object and use arrows to label and determine where all the forces acting upon it will be. then u add up the sum of all forces. if two forces of equal magnitude and opposite directions will cancel each other out.

3 0

3 years ago

Ina Crespo rowed 12 miles down the Habashabee River in 1.5 hours, but the return trip took her 4 hours. Find the rate Ina rows

Alex777 [14]

Answer:

vₓ = 5.5 mile / h and $v_{y}$ = 2.5 mile/h

Explanation:

This is a problem of adding vectors, but since the canoe and the river are in the same direction, we can make an ordinary sum, write the equations for each situation

I rowed down the river. In this case the speed of the canoe and the river are in the same direction, consequently, they add up

(vₓ + $v_{y}$ ) = d / 1.5

I rowed up. In this case the canoe and the river have reversed directions

(vₓ- $v_{y}$ ) = d / 4

Feel us two equations with two unknowns,

Let's start by adding the equations

2vₓ = d / 1.5 + d / 4

2vₓ = 12 (4 + 1.5) / 4 1.5

vₓ = 11/2

vₓ = 5.5 mile / h

Let's substitute in the first of the two equations to find the speed of the river

(vₓ + $v_{y}$ ) = d / 1.5

$v_{y}$ = d / 1.5 - vr

$v_{y}$ = 12 / 1.5 -5.5

$v_{y}$ = 8-5.5

$v_{y}$ = 2.5 mile / h

4 0

3 years ago