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Anettt [7]
3 years ago
12

Please help :/ The same motor is used in rockets with different masses. The rockets have different accelerations. According to N

ewton’s second law, how is acceleration expected to change as the rocket mass increases?
Acceleration cannot be predicted based on changes in mass.
As rocket mass increases, acceleration decreases.
As rocket mass increases, acceleration increases.
There are no changes in acceleration, as it would depend on the amount of force. _

Different amounts of force are applied to the same boat over several trials, and the acceleration is measured. In a graph of acceleration versus net force, what does the slope of the graph represent?
the speed of the boat _
the mass of the boat _
the inverse of the mass of the boat _
the inverse of the speed of the boat
Physics
1 answer:
Ymorist [56]3 years ago
7 0

Answer:

1. As rocket mass increases, acceleration decreases.

2. The inverse of the mass of the boat.

Explanation:

1. Newton's second law of motion states;

         F = ma

where F is the force applied, m is the mass and a is the acceleration.

Therefore, increasing the mass of a rocket increases its weight which would reduce its acceleration provided that the force is constant. Thus, as rocket mass increases, acceleration decreases.

2. The slope of the graph can be expressed as;

From Newton's second law,

F = ma

Slope = (Δa) ÷ (ΔF)

Slope = \frac{a}{F}

⇒ \frac{1}{m} = \frac{a}{F}

Therefore, the slope of the graph is the reciprocal of the mass of the boat.

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A submarine is stranded on the bottom of the ocean with its hatch 21.0 m below the surface. calculate the force (in n) needed to
Tanzania [10]

Answer:

28,400 N

Explanation:

Let's start by calculating the pressure that acts on the upper surface of the hatch. It is given by the sum of the atmospheric pressure and the pressure due to the columb of water, which is given by Stevin's law:

p_{top} = p_{atm} + \rho g h=1.013\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(21.0 m)=3.071 \cdot 10^5 Pa

On the lower part of the hatch, there is a pressure equal to

p_{bot}=p_{atm}=1.013\cdot 10^5 Pa

So, the net pressure acting on the hatch is

p=p_{top}-p_{bot}=3.071 \cdot 10^5 Pa - 1.013\cdot 10^5 Pa=2.058 \cdot 10^5 Pa

which acts from above.

The area of the hatch is given by:

A=\pi r^2 = \pi (\frac{0.420 m}{2})^2=0.138 m^2

So, the force needed to open the hatch from the inside is equal to the pressure multiplied by the area of the hatch:

F=pA=(2.058\cdot 10^5 Pa)(0.138 m^2)=28,400 N

8 0
3 years ago
If you have 10.0 g of a substance that decays with a half-life of 14 days, then how much will you have after 42 days?
Nat2105 [25]
The formula for the mass that remains:
m=m_0 \times (\frac{1}{2})^\frac{t}{T}
m₀ - the initial mass, t - time, T - the half-life

m_0=10 \ g \\
T=14 \ d \\
t=42 \ d \\ \\
m=10 \times (\frac{1}{2})^\frac{42}{14}=10 \times (\frac{1}{2})^3=10 \times \frac{1}{8}=10 \times 0.125=1.25

The answer is c. 1.25 g.
6 0
3 years ago
At which angle must a laser beam enter the water for no refraction to occur?
abruzzese [7]

Light that enters the new medium <em>perpendicular to the surface</em> keeps sailing straight through the new medium unrefracted (in the same direction).

Perpendicular to the surface is the "normal" to the surface. So the angle of incidence (angle between the laser and the normal) is zero, and the law of refraction (just like the law of reflection) predicts an angle of zero between the normal and the refracted (or the reflected) beam.

Moral of the story:  If you want your laser to keep going in the same direction after it enters the water, or to bounce back in the same direction it came from when it hits the mirror, then shoot it <em>straight on</em> to the surface, perpendicular to it.

5 0
3 years ago
An object is placed 4.0 cm to the left of a convex lens with a focal length of +8.0 cm . Where is the image of the object?
Serjik [45]

The image of the object is 8cm to the left of the lens (D)

<h3></h3>

What is the image of an object?

The image of an object is said to be the location where light rays from that object intersect with a mirror by reflection.

It is calculated thus:

1÷v = 1÷f - 1÷u

<h3>How to calculate the image of an object</h3>

From the formula

1÷v = 1÷f - 1÷u

<h3>Where </h3>

V = image distance fromthe object

U = object

f = focal length

Substitute the values

1÷v = 1÷8 - 1÷ 4

1÷v = - 1÷8

Make v the subject of formula

v = -8cm

Therefore, the image of the object is 8cm to the left of the lens (D)

Learn more on focal length here:

brainly.com/question/25779311

#SPJ1

6 0
2 years ago
A spherical tank with radius 4 m is half full of a liquid that has a density of 900 kg/m3. The tank has a 1 m spout at the top.
kompoz [17]

The answer & explanation for this question is given in the attachment below.

7 0
3 years ago
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