All categories

3 years ago

A racecar accelerates from rest at 6.5 m/s2 for 4.1 s. How fast will it be going at the end of that time?

Physics

1 answer:

Pie3 years ago

6 0

Answer:

The final velocity of the car is 26.65 m/s.

Explanation:

Given;

acceleration of the racecar, a = 6.5 m/s²

initial velocity of the car, u = 0

time of motion, t = 4.1 s

The final velocity of the car is given by;

v = u + at

where;

v is the final velocity of the car

suvstitute the givens

v = 0 + (6.5)(4.1)

v = 26.65 m/s.

Therefore, the final velocity of the car is 26.65 m/s.

You might be interested in

Sound waves need a medium to move through in order to keep moving. Through which of the following would sound travel the slowest

Troyanec [42]

Answer:A air

Explanation:Of the three mediums (gas, liquid, and solid) sound waves travel the slowest through gases, faster through liquids, and fastest through solids. Temperature also affects the speed of sound.

Hope this helps you out ツ

8 0

3 years ago

Read 2 more answers

A submarine emerges 1/9 of its volume when it partially floats on the sea surface. For make it completely submerge it is necessa

AleksAgata [21]

Answer:

4,524,660 N

Explanation:

Assuming the submarine's density is uniform, 1/9th of the submarine's mass is equal to the mass of the displaced water.

m/9 = (1026 kg/m³) (50 m³)

m = 461,700 kg

mg = 4,524,660 N

3 0

3 years ago

A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The

Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}$

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

$\sum {F = ma}$

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

$F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg$

Here, ${\mu _{\rm{s}}}$ is the coefficient of static friction, mm is mass of the object, and $g$ is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass ${m_1}$

Substitute for in the expression of maximum static friction ${F_{\rm{f}}} = {\mu _{\rm{s}}}mg$

${F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g$

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute for , [/tex]{m_1}[/tex] for in the equation .

${F_{\rm{f}}} = {m_1}a$

Substitute ${\mu _{\rm{s}}}{m_1}g$ for ${F_{\rm{f}}}$ in the equation ${F_{\rm{f}}} = {m_1}a$

${\mu _{\rm{s}}}{m_1}g = {m_1}a$

Rearrange for a.

$a = {\mu _{\rm{s}}}g$

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right)a$

Substitute for in the above equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

5 0

3 years ago

The stiffness of a particular spring is 42 N/m. One end of the spring is attached to a wall. A force of 2 N is required to hold

Liula [17]

Answer:

$L_{o}=0.1224m$

Explanation:

Given data

Force F=2 N

Length L=17 cm = 0.17 m

Spring Constant k=42 N/m

To find

Relaxed length of the spring

Solution

From Hooke's Law we know that

$F_{spring}=k_{s}|s|\\F_{spring}=k_{s}(L-L_{o})\\ 2N=(42N/m)(0.17m-L_{o})\\2=7.14-42L_{o}\\-42L_{o}=2-7.14\\42L_{o}=5.14\\L_{o}=(5.14/42)\\L_{o}=0.1224m$

6 0

4 years ago

How does a bimetallic strip work in a sinple alarm system

tatiyna

The bimetallic strip in a fire alarm is made of two metals with different expansion rates bonded together to form one piece of metal. Typically, the low-expansion side is made of a nickel-iron alloy called Invar, while the high-expansion side is an alloy of copper or nickel. The strip is electrically energized with a low-voltage current. When the strip is heated by fire, the high-expansion side bends the strip toward an electrical contact. When the strip touches that contact, it completes a circuit that triggers the alarm to sound. The width of the gap between the contacts determines the temperature that will set off the alarm.

7 0

3 years ago

Read 2 more answers