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3 years ago

More questions in the image.

Physics

2 answers:

tensa zangetsu [6.8K]3 years ago

3 0

Answer:

2 True

Explanation: The arrows on a motion map should point in the direction of motion.

nignag [31]3 years ago

3 0

Answer:

1) in a motion map the arrows point in which direction the object is moving, but if there are other arrows in the map (like forces and such) those may point in other direction, so if only movement (velocity) arrows are depicted, this is true.

2) True, as "longer" is the arrow, bigger is the magnitude represented (again, arrows may depict other things than velocity)

3) If only velocity arrows are depicted, this is true, because the velocity is zero, then there is no arrow of velocity.

If other arrows are depicted, like acceleration or force, those may be in there in regards that the object is not moving.

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An 8.00 kg mass is at rest on an incline, the angle of the incline is 32.0°. The incline has a rough surface (μs = 0.375 and μk

neonofarm [45]

Answer:

F= 102,378 N

Explanation:

Attached files (photos).

5 0

4 years ago

A new ride being built at an amusement park includes a vertical drop of 126.5 meters. Starting from rest, the ride vertically dr

yan [13]

Answer:

49.79 m/s

Explanation:

Given:

Initial velocity of the roller coaster is, $u=0\ m/s$

Vertical drop or the displacement of the roller coaster is, $y=-126.5\ m$

The displacement is negative as the motion is in downward direction.

Now, as the motion is in vertical direction only, the acceleration of the roller coaster will be due to gravity acting in the downward direction.

So, the acceleration of the roller coaster is, $a=g=-9.8\ m/s^2$

Now, using the following equation of motion:

$v^2=u^2+2ay$

Where, 'v' is the velocity of the roller coaster at the bottom.

Plug in all the given values and solve for 'v'. This gives,

$v^2=0^2+2(-9.8)(-126.5)\\\\v^2=2479.4\\\\v=\sqrt{2479.4}\\\\v=49.79\ m/s$

Therefore, the speed of the roller coaster at the bottom of the drop is 49.79 m/s.

7 0

3 years ago

A proton moves through an electric potential created by a number of source charges. Its speed is 2.5 x 105 m/s at a point where

Kaylis [27]

Answer:

Proton’s speed, a short time later when it reaches a point of lower potential is 1.4 x 10⁵ m/s

Explanation:

Given;

initial speed of proton, u = 2.5 x 10⁵ m/s

initial potential, V = 1500 V

mass of proton = 1.67 x 10⁻²⁷ kg

Work done, W = eV= ΔK.E = ¹/₂mu²

eV = ¹/₂mu² (J)

where;

e is the charge of the proton in coulombs

V is the electric potential in volts

m is the mass of the proton in kg

u is the speed of the proton in m/s

$m =\frac{2eV_1}{u_1{^2}} = \frac{2eV_2}{u_2{^2}} = \frac{V_1}{u_1{^2}}} =\frac{V_2}{u_2{^2}}$

$\frac{V_1}{u_1{^2}}} =\frac{V_2}{u_2{^2}} = \frac{1500}{(2.5*10^5)^2} = \frac{500}{u_2{^2}} \\\\u_2{^2} =\frac{500*(2.5*10^5)^2}{1500} = 0.333*6.25*10^{10}\\\\u_2 = \sqrt{0.333*6.25*10^{10}} =1.4 *10^5 \ m/s$

Therefore, proton’s speed a short time later when it reaches a point of lower potential is 1.4 x 10⁵ m/s

4 0

3 years ago

What current, in units of milliAmperes, would produce a potential difference of 4.5

Bess [88]

Answer;

<u>= 56.96 mA</u>

Explanation;

From the Ohm's law;

V = IR, where V is the potential difference in volts, I is the electric current in amperes and R is the resistance in ohms.

V = 4.5 volts, R = 79 Ω

Making I the subject of the formula we get;

I =V/R

= 4.5 /79

= 0.05696 A

But, 1 A = 1000 mA

Thus; 0.05696 × 1000

<u>= 56.96 mA</u>

5 0

3 years ago

A child of mass 25 kg is skating fast, +10 m/s, and tries to get revenge by colliding with the 60 kg adult who is sitting still.

tankabanditka [31]

Answer: -4.4 m/s

Explanation:

This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :