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3 years ago

how does the size of objects impact the pull of gravity between Earth and a baseball thrown into the air

2 answers:

3 0

it's how much it weighs and how much force is pushing on it like a egg if i drop it the weigh can cause it to break and how much force the gravity is pushing on it.

attashe74 [19]3 years ago

3 0

The size of an object tossed into the air doesn't matter. The gravitational force on it depends on its mass not its size. Gravity exerts more force on a small rock then on a large styrofoam wiffle ball.

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A large fake cookie sliding on a horizontal surface is attached to one end of a horizontal spring with spring constant k = 440 N

irinina [24]

Answer:

a) 0.275 m b) 13.6 J

Explanation:

In absence of friction, the energy is exchanged between the spring (potential energy) and the cookie (kinetic energy), so at any point, the sum of both energies must be the same:

E = ½ kx2 + ½ mv2

If we take as initial state, the instant when the cookie is passing through the spring’s equilibrium position, all the energy is kinetic, and we know that is equal to 20.0 J.

After sliding to the right, while is being acted on by a friction force, it came momentarily at rest. At this point, the initial kinetic energy, has become potential elastic energy, in part, and in thermal energy also, represented by the work done by the friction force.

So, for this state, we can say the following:

Ki = Uf + Eth = ½* k*d2 + Ff*d

20.0J = ½ *440 N/m* d2 + 11.0 *d, where d is the compressed length of the spring, which is equal to the distance travelled by the cookie before coming momentarily at rest.

We have a quadratic equation, that, after simplifying terms, can be solved as follows, applying the quadratic formula:

d = -0.05/2 +/- √0.090625 = -0.025 +/- 0.3 = 0.275 m (we take the positive root)

b) If we take as our new initial status the moment at which the spring is compressed, and the cookie is at rest, all the energy is potential:

E = Ui = 1/2 k d²

In this case, d is the same value that we got in a), i.e., 0.275 m (as the distance travelled by the cookie after going through the equilibrium point is the same length that the spring have been compressed).

E= 1/2 440 N/m . (0.275)m² = 16.6 J

When the cookie passes again through the equilibrium position, the energy will be in part kinetic, and in part, it will have become thermal energy again.

So, we can write the following equation:

Kf = Ui - Ff.d = 16.6 J - 11.0 (0.275) m = 16.6 J - 3.03 J = 13.6 J

3 0

3 years ago

You are designing a generator with a maximum emf 8.0 V. If the generator coil has 200 turns and a cross-sectional area of 0.030

shutvik [7]

Answer:

7.1 Hz

Explanation:

In a generator, the maximum induced emf is given by

$\epsilon= 2\pi NAB f$

where

N is the number of turns in the coil

A is the area of the coil

B is the magnetic field strength

f is the frequency

In this problem, we have

N = 200

$A=0.030 m^2$

$\epsilon=8.0 V$

B = 0.030 T

So we can re-arrange the equation to find the frequency of the generator:

$f=\frac{\epsilon}{2\pi NAB}=\frac{8.0 V}{2\pi (200)(0.030 m^2)(0.030 T)}=7.1 Hz$

4 0

3 years ago

Read 2 more answers

Which direction does air circulate into low pressure zones in the northern and southern hemispheres

photoshop1234 [79]

Warm air goes up cool air goes down

5 0

3 years ago

Read 2 more answers

Peg P is driven by the forked link OA along the path described by r = eu, where r is in meters. When u = p4 rad, the link has an

8_murik_8 [283]

Answer:

The transverse component of acceleration is 26.32 $m/s^2$ where as radial the component of acceleration is 8.77 $m/s^2$

Explanation:

As per the given data

u=π/4 rad

ω=u'=2 rad/s

α=u''=4 rad/s

$r=e^u$

So the transverse component of acceleration are given as

$a_{\theta}=(ru''+2r'u')\\$

Here

$r=e^u\\r=e^{\pi/4}\\r=2.1932 m$

$r'=e^u.u'\\r'=2.1932 \times 2\\r'=4.3864 m$

$a_{\theta}=(ru''+2r'u')\\a_{\theta}=(2.1932\times 4+2\times 4.3864 \times 2)\\a_{\theta}=26.32 m/s\\$

The transverse component of acceleration is 26.32 $m/s^2$

The radial component is given as

$a_r=r''-r\theta'^2$

Here

$r''=e^u.u'^2+e^u u''\\r''=2.1932 \times (2)^2+2.1932\times 4\\r''=17.5456 m$

$a_r=r''-ru'^2\\a_r=17.5456-2.1932\times (2)^2\\a_r=8.7728 m/s^2$

The radial component of acceleration is 8.77 $m/s^2$

6 0

3 years ago

At t=0, a block A of mass 8 kg and block B of mass 16 kg are both at position x=0 . Block A is at rest, and block B is moving at

love history [14]

The center of mass of the two objects is the average position of the parts of the two object system

The center of mass of block A, and block B after displacement of block B is at 20 m from block A

Reason:

The given parameters are;

The position of block A and block B at t = 0 is x = 0

The mass of block A, m₁ = 8 kg

Mass of block B, m₂ = 16 kg

Speed of block A = 0 m/s

Speed of block B, v₂ = 10 m/s

Location of the center of mass of the two object at t = 3 s; Required

Solution;

The location of block A, after 3 s is x₁ = 0 (block A is at rest)

The location of block B, = v₂ × t

The location of block B, after 3 s is x₂ = 10 m/s × 3 s = 30 m

The center of mass of two masses are given as follows;

$x_{cm} = \dfrac{m_1 \cdot x_1 +m_2\cdot x_2}{m_1 + m_2}$

$x_{cm} = \dfrac{8 \times0 + 16 \times 30}{8 + 16} = 20$

The center of mass of the two objects is at at the position x = 20 m (from block A)

Learn more about the center of mass here:

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4 0

2 years ago