All categories

3 years ago

Car has negative acceleration. What information can you infer from this?

Physics

2 answers:

postnew [5]3 years ago

5 0

Answer:

Its D.

Explanation:

I took the quiz

e-lub [12.9K]3 years ago

4 0

The answer is D. When it is negative it means going down or back. In this case the acceleration is going down. So that means it's slowing down. Hope this helps, rate me please.

You might be interested in

Consider two celestial objects with masses m1 and m2 with a separation distance between their centers r. If the first mass m1 we

Scilla [17]

The new magnitude of the force of attraction will be 6 times the original force of attraction

<h3>How to determine the initial force </h3>

Mass 1 = m₁
Mass 2 = m₂
Gravitational constant = G
Distance apart = r
Initial force (F₁) = ?

F = Gm₁m₂ / r²

F₁ = Gm₁m₂ / r²

<h3>How to determine the new force </h3>

Mass 1 = 2m₁
Mass 2 = 3m₂
Gravitational constant = G
Distance apart (r) = r
New force (F₂) =?

F = Gm₁m₂ / r²

F₂ = G × 2m₁ × 3m₂ / r²

F₂ = 6Gm₁m₂ / r²

But

F₁ = Gm₁m₂ / r²

Therefore

F₂ = 6Gm₁m₂ / r²

F₂ = 6F₁

Thus, the new magnitude of the force of attraction will be 6 times the original force of attraction

Learn more about gravitational force:

brainly.com/question/21500344

#SPJ1

6 0

2 years ago

NEED HELP ASAP WILL MARK BRAINLIEST.

-Dominant- [34]

G is the answer for apex vs / Chehhhh

3 0

3 years ago

A large container contains a large amount of water. A hole is drilled on the wall of the container, at a vertical distance h = 0

barxatty [35]

Answer:

Velocity = 3.25[m/s]

Explanation:

This problem can be solved if we use the Bernoulli equation: In the attached image we can see the conditions of the water inside the container.

In point 1, (surface of the water) we have the atmospheric pressure and at point 2 the water is coming out also at atmospheric pressure, therefore this members in the Bernoulli equation could be cancelled.

The velocity in the point 1 is zero because we have this conditional statement "The water surface drops very slowly and its speed is approximately zero"

h2 is located at point 2 and it will be zero.

$(P_{1} +\frac{v_{1}^{2} }{2g} +h_{1} )=(P_{2} +\frac{v_{2}^{2} }{2g} +h_{2} )\\P_{1} =P_{2} \\v_{1}=0\\h_{2} =0\\v_{2}=\sqrt{0.54*9.81*2}\\v_{2}=3.25[m/s]$

4 0

3 years ago

A thin uniform rod of mass M and length L is bent at its center so that the two segments are now perpendicular to each other. Fi

Tatiana [17]

Answer:

(a) I_A=1/12ML²

(b) I_B=1/3ML²

Explanation:

We know that the moment of inertia of a rod of mass M and lenght L about its center is 1/12ML².

(a) If the rod is bent exactly at its center, the distance from every point of the rod to the axis doesn't change. Since the moment of inertia depends on the distance of every mass to this axis, the moment of inertia remains the same. In other words, I_A=1/12ML².

(b) The two ends and the point where the two segments meet form an isorrectangle triangle. So the distance between the ends d can be calculated using the Pythagorean Theorem:

$d=\sqrt{(\frac{1}{2}L) ^{2}+(\frac{1}{2}L) ^{2} } =\sqrt{\frac{1}{2}L^{2} } =\frac{1}{\sqrt{2} } L=\frac{\sqrt{2} }{2} L$

Next, the point where the two segments meet, the midpoint of the line connecting the two ends of the rod, and an end of the rod form another rectangle triangle, so we can calculate the distance between the two axis x using Pythagorean Theorem again:

$x=\sqrt{(\frac{1}{2}L)^{2}-(\frac{\sqrt{2}}{4}L) ^{2} } =\sqrt{\frac{1}{8} L^{2} } =\frac{1}{2\sqrt{2}} L=\frac{\sqrt{2}}{4} L$

Finally, using the Parallel Axis Theorem, we calculate I_B:

$I_B=I_A+Mx^{2} \\\\I_B=\frac{1}{12} ML^{2} +\frac{1}{4} ML^{2} =\frac{1}{3} ML^{2}$

5 0

3 years ago

A bottle dropped from a balloon reaches the ground in 20 s. determine the height of the balloon if (a) it was at rest in the air

romanna [79]

<span>a) 1960 m b) 960 m Assumptions. 1. Ignore air resistance. 2. Gravity is 9.80 m/s^2 For the situation where the balloon was stationary, the equation for the distance the bottle fell is d = 1/2 AT^2 d = 1/2 9.80 m/s^2 (20s)^2 d = 4.9 m/s^2 * 400 s^2 d = 4.9 * 400 m d = 1960 m For situation b, the equation is quite similar except we need to account for the initial velocity of the bottle. We can either assume that the acceleration for gravity is negative, or that the initial velocity is negative. We just need to make certain that the two effects (falling due to acceleration from gravity) and (climbing due to initial acceleration) counteract each other. So the formula becomes d = 1/2 9.80 m/s^2 (20s)^2 - 50 m/s * T d = 1/2 9.80 m/s^2 (20s)^2 - 50m/s *20s d = 4.9 m/s^2 * 400 s^2 - 1000 m d = 4.9 * 400 m - 1000 m d = 1960 m - 1000 m d = 960 m</span>

6 0

3 years ago