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3 years ago

We can catch a rolling ball but not a flying bullet?? give reason

2 answers:

6 0

A bullet will be moving much faster than a rolling ball. Even seeing a bullet in flight requires it to be extremely low velocity.

Triss [41]3 years ago

3 0

Answer:

yeah this statement is tru

Explanation:

it is because the speed of the bullet is more than the speed of rolling ball .so from this reason we cannot catch a bullet.

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Consider a container of oxygen gas at a temperature of 23°C that is 1.00 m tall. Compare the gravitational potential energy of a

Sergio039 [100]

Answer:

Yes, it is reasonable to neglect it.

Explanation:

Hello,

In this case, a single molecule of oxygen weights 32 g (diatomic oxygen) thus, the mass of kilograms is (consider Avogadro's number):

$m=1molec*\frac{1mol}{6.022x10^{23}molec} *\frac{32g}{1mol}*\frac{1kg}{1000g}=5.31x10^{-26}kg$

After that, we compute the potential energy 1.00 m above the reference point:

$U=mhg=5.31x10^{-26}kg*1.00m*9.8\frac{m}{s^2}=5.2x10^{-25}J$

Then, we compute the average kinetic energy at the specified temperature:

$K=\frac{3}{2}\frac{R}{Na}T$

Whereas $N_A$ stands for the Avogadro's number for which we have:

$K=\frac{3}{2} \frac{8.314\frac{J}{mol*K}}{6.022x10^{23}/mol}*(23+273)K\\ \\K=6.13x10^{-21}J$

In such a way, since the average kinetic energy energy is about 12000 times higher than the potential energy, it turns out reasonable to neglect the potential energy.

Regards.

8 0

3 years ago

7. Two people are pushing a 40.0kg table across the floor. Person 1 pushes with a force of 490N

artcher [175]

Answer:

20.4 $m/s^{2}$

Explanation:

To start doing this problem, first draw a free body diagram of the table. My teacher always tells us to do this, and I find that it is very helpful. I have attached a free body diagram to this answer- take a look at it.

First, let us see if Net force = MA. To do that, we need to determine whether the object is at equilibrium horizontally. For an object to be at equilibrium, it either needs to be moving at a constant velocity or not moving at all. Also, if an object is at equilibrium, there will not be any acceleration. But we know that there IS acceleration horizontally, so it cannot be in equilibrium. If it is not in equilibrium, we can use the formula ∑F= ma.

Let us determine the net force. Since the object is moving horizontally, we can ignore the weight and normal force, because they are vertical forces. The only horizontal forces we need to worry about are the applied force and force of friction.

Applied force = 1055 N (490 + 565)

Friction force= Unknown

To find the friction force, use the kinetic friction formula, Friction = μkN

μk is the coefficient, which the problem includes- it is 0.613.

N is the normal force, which we have to find.

*To find the normal force, we have to determine if the object is at equilibrium VERTICALLY. Since it has no acceleration vertically (it's not moving up/down), it is at equilibrium. Now, when an object is at equilibrium in one direction, it means that all the forces in that direction are equal. What are our vertical forces? Weight (mg) and Normal force (N). So it means that the Normal force is equal to the Weight.

Weight = mg = (40)(9.8) = 392 N

Normal force = 392 N

Now, plug it back into the formula (μkN): (0.613)(392) = 240.296 N

Friction = 240.296 N

Now that we know the friction, we can find the horizontal net force. Just subtract the friction force, 240.296 from the applied force, 1055 N

Horizontal Net Force: 814.704 N

Now that we know the net force, plug in the numbers for the formula

∑F= ma.

814.704 = (40.0)(a)

*Divide on both sides)

a = 20.3676 m/s^2

Round it to 3 significant figures, to get:

20.4 $m/s^{2}$

7 0

3 years ago

A bicyclist with a mass of 50 kg is traveling at a rate of 30 m/s. It accelerates to a rate of 50 m/s in 5 seconds. What is the

balandron [24]

Answer:

F=m*(v^2/r)

F=82*(8^2/30)

F=174.9N

Explanation:

brainlest pls

8 0

3 years ago

What is the magnitude of the torque that the axle must apply to prevent the disk from rotating?

mihalych1998 [28]

The required torque at the axle, is given by the difference between the

moments of the applied forces.

The torque required is 19.62 N·m counterclockwise

Reasons:

The given parameters are;

Mass of the disk, m = 5.0 kg

Location of the axle = Half the radius of the disk

Diameter of the disk, D = 40 cm = 0.4 m

Applied mass, 0.1 m from the axle = 15 kg

Applied mass, 0.3 m from the axle = 10 kg

Required:

Magnitude of torque at the axle that prevent the disk from rotating

Solution:

Torque needed = Clockwise moment - Counterclockwise moment

Clockwise moment = (10 kg × 0.3 m + 5 kg × 0.1 m) × 9.81 m/s² = 34.335 N·m

Counterclockwise moment = 15 kg × 0.1 m × 9.81 m/s² = 14.715 N·m

τ + Counterclockwise moment = Clockwise moment

τ + 14.715 N·m = 34.335 N·m

Torque required, τ = 34.335 N·m - 14.715 N·m = 19.62 N·m

Torque required, τ = 19.62 N·m counterclockwise

Learn more here:

brainly.com/question/19044661

brainly.com/question/19247046

The probable question drawing obtained from a similar question online is attached

7 0

2 years ago

The arrows on the diagram show the ocean floor spreading from the Redge what three kinds of evidence scientists have found suppo

Blizzard [7]

Edit: You do mean Ridge?

Rocks near Mid-Ocean Ridge are younger than rocks near the trenches.

Seismic data shows oceanic crust is sinking into the mantle at the trenches.

Matching bands of magnetic rock are found on either side of the Ridge. Earth's magnetic fields change these bands over time.

5 0

3 years ago