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Paha777 [63]
3 years ago
11

A juggler throws a bowling pin straight up in the air. After the pin leaves his hand and while it is in the air, which statement

is true?(a) The velocity of the pin is always in the same direction as its acceleration.(b) The velocity of the pin is never in the same direction as its acceleration.(c) The acceleration of the pin is zero.(d) The velocity of the pin is opposite its acceleration on the way up. (e) The velocity of the pin is in the same direction as its acceleration on the way up.
Physics
1 answer:
WINSTONCH [101]3 years ago
8 0

Answer:

The velocity of the pin is opposite its acceleration on the way up.

(d) option is correct.

Explanation:

when the juggler throws a bowling pin straight in the air, the acceleration working on the pin is in the downward direction due to the gravitational force of the earth.

According to Newton's Universal Law of Gravitation

''The gravitational force is a force that attracts any objects with mass''

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A centrifuge is used to test space pilots. The centrifuge spins with a centripetal acceleration of 3.04g. If the length of the a
uranmaximum [27]

Answer:

approx.= 25\frac{m}{s}

Explanation:

Centripetal acceleration (a) is defined as the square of an object's velocity (V^2) divided by the distance of the object from it's point/axis of revolution (r). So:

a=\frac{V^{2} }{r}

which allows us to solve for the velocity:

V=\sqrt{ar}\\ Since: a=3.04g=(3.04)(9.81),r=21;\\V=\sqrt{(3.04)(9.81)(21)} =25.02...

3 0
3 years ago
Two blocks A and B with mA = 2.2 kg and mB = 0.84 kg are connected by a string of negligible mass. They rest on a frictionless h
madreJ [45]

Answer:

(a) a = 1.875 m/s²

(b)T = 1.575 N

(c)T increase

Explanation:

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the blocks on the horizontal surface and the y-axis in the direction perpendicular to it.

Forces acting on the block  A

WA: Weight of of the A block : In vertical direction  downaward (-y)

NA : Normal force of the A block :In vertical direction  upaward (+y)

F= 5.7 N  In in the direction parallel to the movement of the blocks (+x)

T :  tension of the string: In in the direction (-x)

Forces acting on the block  B

WB: Weight of the B block: In vertical direction  downaward (-y)

NB : Normal force of the B block :In vertical direction  upaward (+y)

T :  tension of the string: In in the direction (+x)

Data

mA = 2.2 kg

mB = 0.84 kg

(a) Magnitude of the acceleration  of the blocks

Newton's second law to B block:

∑Fx = m*a

T = (0.84)*a Equation (1)

Newton's second law to A block:

∑Fx = m*a

5.7 - T = (2.2)*a  We replace T of the Equation (1)

5.7 - (0.84)*a = (2.2)*a  

5.7 = (2.2)*a + (0.84)*a  Equation (2)

5.7 =(3.04)*a

a =5.7 / (3.04)

a = 1.875 m/s²

(b)Tension (in N) in the string connecting the two blocks

We replace data in the  Equation (1)

T = (0.84)*a

T = (0.84)*(1.875)

T = 1.575 N

(c) How will the tension in the string be affected if mA is decreased?

We observed Equation (2)

5.7 = (2.2)*a + (0.84)*a

5.7 = (mA)*a + (0.84)*a

5.7 =a*( mA+ 0.84)

if mA decrease , then, the acceleration increase and T  increase

4 0
3 years ago
What are the major steps of solar system formation in the correct order?
Gala2k [10]

Answer:(1) Pre-solar nebula

(2). Planet formation

Explanation:

Nebular hypothesis says that the Solar System formed from the gravitational collapse of a fragment of a giant molecular cloud.

One of the collapsing fragments called the pre-solar nebula formed what became the Solar System.

The planet can also be formed by accretion. Accretion is a process in which planets began as dust grains in orbit around the central protostar.

When the terrestrial planets were forming, they remained immersed in a disk of gas and dust.

4 0
3 years ago
If an electric wire is allowed to produce a magnetic field no larger than that of the earth (0.50×10−4t) at a distance of 12 cm
Studentka2010 [4]
The magnetic field generated by a wire carrying a current I is:
B(r) =  \frac{\mu_0 I}{2 \pi r}
where r is the distance at which the magnetic field is measured, and \mu_0 = 4 \pi \cdot 10^{-7} NA^{-2} is the magnetic permeability in vacuum.

The problem says that the magnetic field at a distance r=12 cm=0.12 m from the wire must be no larger than B=0.5 \cdot 10^{-4}T. Substituting these values, we can find the maximum value of the current I that the wire can carry:
I= \frac{2 \pi r B}{\mu _0} = \frac{2 \pi (0.12 m)(0.5 \cdot 10^{-4}T)}{ 4 \pi \cdot 10^{-7} NA^{-2}}= 30 A
4 0
3 years ago
Assume that in 2010 the United States will need 2.0×1012 watts of electric power produced by thousands of 1000 MW power plants.
Alex73 [517]

Answer:

1752.14 tonnes per year.

Explanation:

To solve this exercise it is necessary to apply the concepts related to power consumption and power production.

By conservation of energy we know that:

\dot{P} = \bar{P}

Where,

\dot{P} = Production of Power

\bar{P} = Consumption of power

Where the production of power would be,

\dot{P} = m \dot{E}\eta

Where,

m = Total mass required

\dot{E} = Energy per Kilogram

\eta =Efficiency

The problem gives us the aforementioned values under a production efficiency of 45%, that is,

\dot{P} = \bar{P}

m \dot{E}\eta = \bar{P}

Replacing the values we have,

m(8*10^13)(0.45) = 2*10^{12}

Solving for m,

m = \frac{ 2*10^{12}}{(8*10^13)(0.45)}

m = 0.0556 \frac{kg}{s}

We have the mass in kilograms and the time in seconds, we need to transform this to tons per year, then,

m = 0.556\frac{kg}{s}*(\frac{3.1536*10^7s}{1year})(\frac{1ton}{1000kg})

m = 1752.14tonnes per year.

8 0
3 years ago
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