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Marizza181 [45]
3 years ago
11

The speed of an object changes only when it is acted on by an unbalanced force.

Physics
1 answer:
Nikitich [7]3 years ago
6 0

Answer:

If an object has a net force acting on it, it will accelerate. The object will speed up, slow down or change direction. An unbalanced force (net force) acting on an object changes its speed and/or direction of motion. An unbalanced force is an unopposed force that causes a change in motion.

Explanation:

I hope this helps you out and if your feeling generous plz mark brainliest it helps me a lot thank you:)

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A cork shoots out of a champagne bottle at an angle of 40.0 above the horizontal. If the cork travels a horizontal distance of 1
ruslelena [56]

Answer:

The initial speed of the cork was 1.57 m/s.

Explanation:

Hi there!

The equation of the horizontal position of the cork in function of time is the following:

x = x0 + v0 · t · cos θ

Where:

x = horizontal position at time t.

x0 = initial horizontal position.

v0 = initial speed of the cork.

t = time.

θ = launching angle.

If we place the origin of the frame of reference at the launching point, then x0 = 0.

We know that at t = 1.25 s, x = 1.50 m. We also know the launching angle so we can solve the equation of horizontal position for the initial speed, v0:

x = v0 · t · cos θ

x / t · cos θ = v0

v0 = 1.50 m / (1.25 s · cos (40.0°)

v0 = 1.57 m/s

The initial speed of the cork was 1.57 m/s.

4 0
3 years ago
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A bucket that has a mass of 20 kg when filled with sand needs to be lifted to the top of a 15 meter tall building. You have a ro
prohojiy [21]

Answer:

work done lifting the bucket (sand and rope) to the top of the building,

W=67.46 Nm

Explanation:

in this question we have given

mass of bucket=20kg

mass of rope=.2\frac{kg}{m}

height of building= 15 meter

We have to find the work done lifting the bucket (sand and rope) to the building =work done in lifting the rope + work done in lifting the sand

work done in lifting the rope is given as,W_{1}=Force \times displacement

=\int\limits^{15}_0 {.2x} \, dx ..............(1)

=.1\times 15^2

=22.5 Nm

work done in lifting the sand is given as,W_{2}=Force \times displacement

W_{2}=\int\limits^{15}_0 F \, dx.................(2)

Here,

F=mx+c

here,

c=20-18

c=2

m=\frac{20-18}{15-0}

m=.133

Therefore,

F=.133x+2

Put value of F in equation 2

W_{2}=\int\limits^{15}_0 (.133x+2) \, dx

W_{2}=.133 \times 112.5+2\times15\\W_{2}=14.96+30\\W_{2}=44.96 Nm

Therefore,

work done lifting the bucket (sand and rope) to the top of the building,W=W_{1}+W_{2}

W=22.5 Nm+44.96 Nm

W=67.46 Nm

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3 years ago
A waitperson carrying a tray with a platter on it tips the tray at an angle of 12° below the horizontal. If the gravitational fo
Nadusha1986 [10]

Answer:

The answer is 1.0 N

Explanation:

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Now this gravitational force has two component i.e.

5\sin \theta  is parallel to the tray =1.039 N

5\cos \theta  is perpendicular to the tray =4.890 N

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