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Juliette [100K]
3 years ago
8

The gravitational strength at the poles is greater than the gravitational strength at the equator. What will happen to an object

when it moves from the poles to the equator? Group of answer choices A. Its mass will increase. B. Its mass will decrease. C. Its weight will increase. D. Its weight will decrease.
Physics
1 answer:
geniusboy [140]3 years ago
3 0

Answer:

C

Explanation:

Because this same question was on my test last week and I got it correct

You might be interested in
Do all metals stick to magnets? Give an example. I NEED HELP FASTT
ludmilkaskok [199]

Answer:

No

Explanation:

Not all metals stick to magnets. Like aluminum. if you were to stick a magnet on to an aluminum it would fall off.

6 0
2 years ago
A ball is thrown from a rooftop with an initial downward velocity of magnitude vo = 2.9 m/s. The rooftop is a distance above the
Step2247 [10]

Answer:

a) The velocity of the ball when it hits the ground is -20.5 m/s.

b) To acquire a final velocity of 27.3 m/s, the ball must be thrown from a height of 38 m.

Explanation:

I´ve found the complete question on the web:

<em />

<em>A ball is thrown from a rooftop with an initial downward velocity of magnitude v0=2.9 m/s. The rooftop is a distance above the ground, h= 21 m. In this problem use a coordinate system in which upwards is positive.</em>

<em>(a) Find the vertical component of the velocity with which the ball hits the ground.</em>

<em>(b) If we wanted the ball's final speed to be exactly 27, 3 m/s from what height, h (in meters), would we need to throw it with the same initial velocity?</em>

<em />

The equation of the height and velocity of the ball at any time "t" are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the ball at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity of the ball at a time "t".

First, let´s find the time it takes the ball to reach the ground (the time at which h = 0)

h = h0 + v0 · t + 1/2 · g · t²

0 = 21 m - 2.9 m/s · t - 1/2 · 9.8 m/s² · t²

Solving the quadratic equation using the quadratic formula:

t = 1.8 s  ( the other solution of the quadratic equation is rejected because it is negative).

Now, using the equation of velocity, let´s find the velocity of the ball at

t = 1.8 s:

v = v0 + g · t

v = -2.9 m/s - 9.8 m/s² · 1.8 s

v = -20.5 m/s

The velocity of the ball when it hits the ground is -20.5 m/s.

b) Now we have the final velocity and have to find the initial height. Using the equation of velocity we can obtain the time it takes the ball to acquire that velocity:

v = v0 + g · t

-27.3 m/s = -2.9 m/s - 9.8 m/s² · t

(-27.3 m/s + 2.9 m/s) / (-9.8 m/s²) = t

t = 2.5 s

The ball has to reach the ground in 2.5 s to acquire a velocity of 27.3 m/s.

Using the equation of height, we can obtain the initial height:

h = h0 + v0 · t + 1/2 · g · t²

0 = h0 -2.9 m/s · 2.5 s - 1/2 · 9.8 m/s² · (2.5 s)²

-h0 = -2.9 m/s · 2.5 s - 1/2 · 9.8 m/s² · (2.5 s)²

h0 = 38 m

To acquire a final velocity of 27.3 m/s, the ball must be thrown from a height of 38 m.

6 0
3 years ago
Two resistors, A and B, are connected in series to a 6.0 V battery. A voltmeter connected across resistor A measures a potential
mestny [16]

Answer:

Resistance of resistor A = 6.0 Ω and resistance of resistor B = 3.0 Ω

Explanation:

When the two resistors are in series, let V₁ = voltage in resistor A and R₁ = resistance of resistor A and V₂ = voltage in resistor B and R₂ = resistance of resistor B.

Given that V₁ + V₂ = 6.0 V and V₁ = 4.0 V,

V₂ = 6.0 V - V₁ = 6.0 V - 4.0 V = 2.0 V

Also, let the current in series be I.

So, V₁ = IR₁ and V₂ = IR₂

I = V₁/R₁ and I = V₂/R₂

equating both expressions, we have

V₁/R₁ = V₂/R₂

4.0 V/R₁ = 2.0 V/R₂

dividing through by 2.0 V, we have

2/R₁ = 1/R₂

taking the reciprocal, we have

R₂ = R₁/2

R₁ = 2R₂

From the parallel connection, let V₁ = voltage in resistor A and R₁ = resistance of resistor A and V₂ = voltage in resistor B and R₂ = resistance of resistor B. Since it is parallel, V₁ = V₂ = V = 6.0 V

Also, V₂ = I₂R₂ where I₂ = current in resistor B = 2.0 A and R₂ = resistance of resistor B

So, R₂ = V₂/I₂

= 6.0 V/2.0 A

= 3.0 Ω

R₁ = 2R₂

= 2(3.0 Ω)

= 6.0 Ω

So, resistance of resistor A = 6.0 Ω and resistance of resistor B = 3.0 Ω

6 0
3 years ago
Young's Modulus refers to changes in the a Volume b- Length c- Body layers
Novosadov [1.4K]

Explanation:

Young' modulus is the ratio of normal stress to the longitudinal strain. Mathematically, it is given by :

Normal stress is given by force per unit area. Longitudinal strain is the change in length per unit original length.

The mathematical definition of Young's modulus is given by :

Y=\dfrac{F/A}{\Delta L/L}..........(1)

Where

\Delta L is the change in length

F is the force

A is the area of cross section

So, the Young's modulus refers to the change in length of the object. Hence, the correct option is (b) "length".

8 0
3 years ago
A projectile falls beneath the straight-line path it would follow if there were no gravity. How many meters does it fall below t
maria [59]

Answer:

Explanation:

Suppose v is the initial velocity and \theta is the angle of inclination

distance traveled in vertical direction in t=1 s

When gravity is present

y=vt+\frac{1}{2}at^2

where y=vertical\ distance

a=acceleration

t=time

v=initial\ velocity

here initial velocity is v\sin \theta [/tex] so

y=v\sin \theta \times 1-\frac{1}{2}gt^2

y=v\sin \theta -0.5g

In absence of gravity

y_2=v\sin \theta \times t

y_2=v\sin \theta \times 1

\Delta y=y_2-y=v\sin \theta -v\sin \theta +0.5 g=4.9\ m

         

4 0
3 years ago
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