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eimsori [14]
3 years ago
5

Which is not an example of a scalar? a. 2t/s b. 3kg c. 6.2m north d. -100c

Physics
1 answer:
xxMikexx [17]3 years ago
4 0
C because I’m a teacher
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Consider the same roller coaster. It starts at a height of 40.0 m but once released, it can only reach a height of 25.0 m above
poizon [28]

Answer:

The magnitude of the frictional force between the car and the track is 367.763 N.

Explanation:

The roller coster has an initial gravitational potential energy, which is partially dissipated by friction and final gravitational potential energy is less. According to the Principle of Energy Conservation and Work-Energy Theorem, the motion of roller coster is represented by the following expression:

U_{g,1} = U_{g,2} + W_{dis}

Where:

U_{g,1}, U_{g,2} - Initial and final gravitational potential energy, measured in joules.

W_{dis} - Dissipated work due to friction, measured in joules.

Gravitational potential energy is described by the following formula:

U = m \cdot g \cdot y

Where:

m - Mass, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

y - Height with respect to reference point, measured in meters.

In addition, dissipated work due to friction is:

W_{dis} = f \cdot \Delta s

Where:

f - Friction force, measured in newtons.

\Delta s - Travelled distance, measured in meters.

Now, the energy equation is expanded and frictional force is cleared:

m \cdot g \cdot (y_{1} - y_{2}) = f\cdot \Delta s

f = \frac{m \cdot g \cdot (y_{1}-y_{2})}{\Delta s}

If m = 1000\,kg, g = 9.807\,\frac{m}{s^{2}}, y_{1} = 40\,m, y_{2} = 25\,m and \Delta s = 400\,m, then:

f = \frac{(1000\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (40\,m-25\,m)}{400\,m}

f = 367.763\,N

The magnitude of the frictional force between the car and the track is 367.763 N.

7 0
3 years ago
You drive a beat-up pickup truck along a straight road for 8.4 km at 70 km/h, at which point the truck runs out of gasoline and
andreev551 [17]

Answer:

v_avg = 37 km/h

Explanation:

To find the average velocity in the complete trajectory you use the following formula:

v_{avg}=\frac{v_1+v_2}{2}   ( 1 )

v1: velocity in the first part of the trajectory = 70 km/h

v2: velocity in the second part of the trajectory = ?

You calculate v2 by using the following equation for a motion with constant velocity:

v_2=\frac{2.0km}{30min}*\frac{60min}{1h}=4\frac{km}{h}

you replace the values of v1 and v2 in (1) and you obtain:

v_{avg}=\frac{70km/h+4km/h}{2}=37\frac{km}{h}

hence, the average velocity is 37 km/h

6 0
3 years ago
Read 2 more answers
What is the kinetic energy that the device contains immediately before it hits the ground? [KE = /2*m*v-]
n200080 [17]

Answer:

Kinetic energy : "It is the energy possessed by body by virtue of its motion" .

Explanation:

The mechanical energy is the energy that a body has due to its motion or position .

It is of two types :

  • Potential energy
  • Kinetic energy

Potential energy

"It is the energy that a body has due to its position or state" .

It is given by P.E=mgh

Kinetic energy

"It is the energy that a body has due to its motion" .

The formula is : K.E=1/2mv².

For example

  • A ball is placed on table .
  • At that time it has potential energy .
  • Now,lets say that we push it and it drops down .
  • On doing this ,the potential energy of ball gets converted into kinetic energy(because it comes in motion )  and as soon as it touches the ground it stops and its kinetic energy gets converted into Potential energy again .
6 0
3 years ago
According to the Periodic Table of Elements, Calcium (Ca) has:
RSB [31]

Answer:

A- 20 protons and 20 electrons

Explanation:

8 0
3 years ago
c. A car was running with a velocity of 20m/s. what will be its velocity after 30s if it's acceleration is 5m/s2​
weqwewe [10]

\large \mathfrak{Solution : }

let's use first equation of motion to solve this ;

  • \boxed{ \boxed{ v = u + at}}

  • v = 20 + (5 \times 30)

  • v = 20 + 150

  • v = 170 \:  \: m/s

Velocity after 30 seconds = 170 m/s

4 0
3 years ago
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