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alexgriva [62]
3 years ago
7

What’s the net force of an object being throwing into the air

Physics
1 answer:
timurjin [86]3 years ago
4 0

Answer: forces acting on an object being thrown into the air is gravity and possibly air resistance

Explanation:

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A tug boat pulls a ship with a constant net horizontal force of 5.00•10*3 N and causes the ship to move through a harbor. How mu
Murljashka [212]

The work done on the ship is 1.5\cdot 10^7 J

Explanation:

The work done by a force on an object is given by:

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and of the displacement

In this problem, we have:

F=5.00\cdot 10^3 N (force acting on the ship)

d = 3.00 km = 3000 m (displacement of the ship)

\theta=0^{\circ} (because the force is horizontal, and the displacement is horizontal as well)

Therefore, the work done on the ship is

W=(5.00\cdot 10^3)(3000)(cos 0^{\circ})=1.5\cdot 10^7 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

8 0
3 years ago
I NEED HELP WITH THE LAST QUESTION I’L MARK U AS BRANLIEST
velikii [3]

Answer:

gap junctions,tight junctions,and desmosomes

6 0
3 years ago
Select the correct answer from each drop-down menu.
earnstyle [38]

Answer:

a transverse (sort of a plot of a sine or cosine graph, basically)

b longitudinal

c Electromagnetic (an electric wave and a magnetic wave travelling together at right angles to each other)

Explanation:

7 0
3 years ago
Assume it takes 10 J to stretch a spring 10 cm beyond its natural length. Find the work required (in Joules) to stretch the spri
Lady bird [3.3K]

Answer:

W = 30 J

Explanation:

given,

Work done = 10 J

Stretch of spring, x = 0.1 m

We know,

dW = F .dx

we know, F = k x

\int dW = \int_0^{0.1} k.x dx

W = \int_0^{0.1} k.x dx

W = k[\dfrac{x^2}{2}]_0^{0.1}

10 = k\dfrac{0.1^2}{2}

k = 2000

now, calculating Work done by the spring when it stretched to 0.2 m from 0.1 m.

W = \int_{0.1}^{0.2} 2000 x dx

W = 2000 [\dfrac{x^2}{2}]_{0.1}^{0.2} dx

W = 1000 x 0.03

W = 30 J

Hence, work done is equal to 30 J.

4 0
3 years ago
slader A jet is circling an airport control tower at a distance of 15.9 km. An observer in the tower watches the jet cross in fr
liubo4ka [24]

Answer:

y = 138.96 m

Explanation:

The angle subtended by the moon is the mean of the angle of the arc between the two most extreme points of the moon, we can see that the angle is very small, so we can approximate this arc to a straight line and then use the trigonometric relationships

         sin θ = y / L

where L = 15.9 10³ m and θ = 8.74 10⁻³ rad

          y = L sin θ

          y = 15.9 10³ sin (8.74 10⁻³)

         y = 15.9 10³    0.0087399

         y = 138.96 m

4 0
2 years ago
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