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andrezito [222]
3 years ago
6

How much do a steel BB weight?

Physics
1 answer:
pychu [463]3 years ago
6 0
Each BB 0.349 grams and 5.386 grains. 349 mg & 5.386 grains each steel BBs.
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Are the objects described here in static equilibrium, dynamic equilibrium, or not equilibrium at all? Explain.
Alexandra [31]
Let us examine the given situations one at a time.

Case a. A 200-pound barbell is held over your head.
The barbell is in static equilibrium because it is not moving.
Answer: STATIC EQUILIBRIUM

Case b. A girder is being lifted at a constant speed by a crane.
The girder is moving, but not accelerating. It is in dynamic equilibrium.
Answer: DYNAMIC EQUILIBRIUM

Case c: A jet plane has reached its cruising speed at an altitude.
The plane is moving at cruising speed, but not accelerating. It is in dynamic equilibrium.
Answer: DYNAMIC EQUILIBRIUM

Case d: A box in the back of a truck doesn't slide as the truck stops.
The box does not slide because the frictional force between the box and the floor of the truck balances out the inertial force. The box is in static equilibrium.
Answer: STATIC EQUILIBRIUM
4 0
3 years ago
Read 2 more answers
one of the lady spartans was falling to the ground after dunking the winning basket. At the end of her fall, she was falling 4 m
Tresset [83]

Answer:

15 is the correct answer I.t.

8 0
3 years ago
Question 2 (1 point)
eduard

Answer: Blake may have a decreased chance of disease.

Explanation:

Research has shown that on average, a person who engages in more physical activity will live longer than a person who does not with one reason for this being that engaging in physical activity helps reduce the risk of diseases such as; heart disease, diabetes, and many forms of cancer.

Obesity (risk factor) and high blood pressure are also reduced by engaging in physical activity.

6 0
3 years ago
Who first used the word atom to describe the smallest unit
Firlakuza [10]

Answer: It was Democritus, in fact, who first used the word atomos to describe the smallest possible particles of matter.

Explanation: hope this helped

4 0
3 years ago
A solid conducting sphere of radius 2.00 cm has a charge of 6.88 μC. A conducting spherical shell of inner radius 4.00 cm and ou
zepelin [54]

Explanation:

Given that,

Radius R= 2.00

Charge = 6.88 μC

Inner radius = 4.00 cm

Outer radius  = 5.00 cm

Charge = -2.96 μC

We need to calculate the electric field

Using formula of electric field

E=\dfrac{kq}{r^2}

(a). For, r = 1.00 cm

Here, r<R

So, E = 0

The electric field does not exist inside the sphere.

(b). For, r = 3.00 cm

Here, r >R

The electric field is

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times6.88\times10^{-6}}{(3.00\times10^{-2})^2}

E=6.88\times10^{7}\ N/C

The electric field outside the solid conducting sphere and the direction is towards sphere.

(c). For, r = 4.50 cm

Here, r lies between R₁ and R₂.

So, E = 0

The electric field does not exist inside the conducting material

(d).  For, r = 7.00 cm

The electric field is

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times(-2.96\times10^{-6})}{(7.00\times10^{-2})^2}

E=5.43\times10^{6}\ N/C

The electric field outside the solid conducting sphere and direction is away of solid sphere.

Hence, This is the required solution.

6 0
3 years ago
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