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Amiraneli [1.4K]
3 years ago
13

If you were rolling a toy car down an incline, what would happen if you made the incline steeper? Explain why this happens using

Newton's Laws.
Physics
1 answer:
german3 years ago
7 0

Answer: When you drop the object, this potential energy is converted into kinetic energy, or the energy of motion. So when you roll a ball down a ramp, it has the most potential energy when it is at the top, and this potential energy is converted to both transnational and rotational kinetic energy as it rolls down.

Explanation:

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Using the graph predict how many paper clipsa 7.5 v battery would pick up for both the 25 coil electromagnet and the 50 coil ele
enyata [817]

The graph between the strength of the magnet(number of paper clips picked) and battery is approximately a straight line.

For 25 coil, with increase of 1.5 V battery voltage, the electromagnet picks about 5 more clips. So, for a 7.5 V battery, it would pick about 30 paper clips.

For 50 coil, with increase of 1.5 V battery voltage, the electromagnet picks about 15 more clips. So, for a 7.5 V battery, it would pick about 30 paper clips.

3 0
3 years ago
A car is initially moving at 10.5 m/s and accelerates uniformly to reach a speed of 21.7 m/s in 4.34 s. How far did the car move
Zarrin [17]

Answer:

A. 69.9m

Explanation:

Given parameters:

Initial velocity = 10.5m/s

Final velocity  = 21.7m/s

Time  = 4.34s

Unknown:

Distance traveled = ?

Solution:

Let us first find the acceleration of the car;

  Acceleration  = \frac{v - u}{t}

  v is final velocity

   u is initial velocity

   t is the time

     Acceleration  = \frac{21.7 - 10.5}{4.34}   = 2.58m/s²

Distance traveled;

     V² = U² + 2aS

    21.7² = 10.5² + 2 x 2.58 x S

   360.64 = 2 x 2.58 x S

     S = 69.9m

3 0
3 years ago
in which of the following collisions would you expect the kinetic energy to be conserved? a. a bullet passes through a block of
Scorpion4ik [409]

An elastic collision is one in which the system does not experience a net loss of kinetic energy as a result of the collision. In elastic collisions, momentum and kinetic energy are both conserved.

<h3>Explain about the Elastic Collision?</h3>

A collision between two bodies in physics is referred to as an elastic collision if their combined kinetic energy stays constant. There is no net conversion of kinetic energy into other forms, such as heat, noise, or potential energy, in an ideal, fully elastic collision

An example of an elastic collision is when two balls collide at a pool table. It is an elastic collision when you throw a ball on the ground and it bounces back into your hand because there is no net change in the kinetic energy.

If there is no kinetic energy lost in the impact, the collision is said to be perfectly elastic. A collision is considered to be inelastic if any of the kinetic energy is converted to another kind of energy during the collision.

To learn more about Elastic Collision refer to:

brainly.com/question/7694106

#SPJ4

8 0
1 year ago
Which statement best explains how isotopes can have different masses and still be the same element?
ss7ja [257]

Your list of choices is a very short list, and doesn't include any
correct explanation.

The mass of an atom is roughly the number of protons AND neutrons
in its nucleus, but the element only depends on how many protons are there. 

Different atoms of the same element may have different numbers of
neutrons, so their masses are different.  But they're the same element,
because they all have the same number of protons.

3 0
3 years ago
Steam enters a counterflow heat exchanger operating at steady state at 0.07 MPa with a specific enthalpy of 2431.6 kJ/kg and exi
PIT_PIT [208]

Answer:

Explanation:

Given:

Steam Mass rate, ms = 1.5 kg/min

= 1.5 kg/min × 1 min/60 sec

= 0.025 kg/s

Air Mass rate, ma = 100 kg/min

= 100 kg/min × 1 min/60 sec

= 1.67 kg/s

A.

Extracting the specific enthalpy and temperature values from property table of “Saturated water – Pressure table” which corresponds to temperature at 0.07 MPa.

xf, quality = 0.9.

Tsat = 89.9°C

hf = 376.57 kJ/kg

hfg = 2283.38 kJ/kg

Using the equation for specific enthalpy,

hi = hf + (hfg × xf)

= 376.57 + (2283.38 × 0.9)

= 2431.552 kJ/kg

The specific enthalpy of the outlet, h2 = hf

= 376.57 kJ/kg

B.

Rate of enthalpy (heat exchange), Q = mass rate, ms × change in specific enthalpy

= ms × (hi - h2)

= 0.025 × (2431.552 - 376.57)

= 0.025 × 2055.042

= 51.37455 kW

= 51.38 kW.

5 0
3 years ago
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