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3 years ago

10

Andrew pokes a marble, and the marble rolls down a ramp. The marble moves with speed. Which forces are acting on the marble in t

his situation?

2 answers:

stich3 [128]3 years ago

4 0

The forces are Andrew poking the marble and then gravity pulling the marble downward

Lera25 [3.4K]3 years ago

3 0

Answer:

1. an applied force 2. gravitational 3.friction

Explanation:

Took it on Plato :)))

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4. AN OBJECT INCREASES ITS VELOCITY

Cloud [144]

Answer:

it is accelerating 14 m/s

Explanation:

7 0

3 years ago

The weight of the mass added to the hanger is equal to the extra force on the gas, but what area should we use to calculate the

kari74 [83]

Answer: according to the Avagadro's law, volume is directly propotional to no of moles: VXn

according to the Charles law, volume is directly propotional to temperatue: VXT

according to the Boyle's law, volume is inversely propotional to P: VX1/P

when we combine them we get:

VXnT1/P

V=knT/P

k= R(universal gas constant)

V=RnT/P

PV=nRT

8 0

3 years ago

Find the following answer based on the image.

saul85 [17]

<u>We are given:</u>

Mass of Neptune = 1.03 * 10²⁶ kg

Distance from the center of Neptune (r) = 2.27 * 10⁷

now, computing the value of the acceleration due to gravity (g)

<u>Finding g:</u>

We know the formula:

g = G(mass of planet) / (r)²

g = [6.67 * 10⁻¹¹ * 1.03*10²⁶] / (2.27*10⁷) [since G is 6.67*10⁻¹¹]

g = (6.87 * 10¹⁵) / (5.15 * 10¹⁴)

which can be rewritten as:

g = (6.87 * 10¹⁵ * 10⁻¹⁴) / 5.15

g = (6.87 * 10¹⁵⁻¹⁴) / 5.15

g = (6.87/5.15) * 10

g = 1.34 * 10

g = 13.4 m/s² <em>(approx)</em>

5 0

3 years ago

A cable is 100-m long and has a cross-sectional area of 1.0 mm2. A 1000-N force is applied to stretch the cable. Young's modulus

Blizzard [7]

Answer:

1 m

Explanation:

L = 100 m

A = 1 mm^2 = 1 x 10^-6 m^2

Y = 1 x 10^11 N/m^2

F = 1000 N

Let the cable stretch be ΔL.

By the formula of Young's modulus

$Y=\frac{F\times L}{A\times\Delta L}$

$\Delta L=\frac{F\times L}{A\times\Y}$

$\Delta L=\frac{1000\times 100}{10^{-6}\times10^{11}}$

ΔL = 1 m

Thus, the cable stretches by 1 m.

4 0

3 years ago

In a football game, a receiver is standing still, having just caught a pass. Before he can move, a tackler, running at a velocit

ivolga24 [154]

Answer:

$m_{receiver}=115Kg*3.1/(1.6)-115Kg=107.8Kg$

Explanation:

The football players collide in a completely inelastic collision, in other words they have the same velocity after the collision, this velocity has a magnitude V=1.6m/s.

We need to use the conservation of momentum Law, the total momentum is the same before and after the collision, at the initial point the receiver does not have any speed

$m_{tackler}*v_{tackler}=(m_{tackler}+m_{receiver})V$ (1)

We solve in order to find the receiver mass:

$m_{receiver}={m_{tackler}*v_{tackler}/V}-m_{tackler}$

$m_{receiver}=115Kg*3.1/(1.6)-115Kg=107.8Kg$

5 0

3 years ago