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dolphi86 [110]
2 years ago
5

A car advertisement states that a certain car can accelerate from rest to

Physics
2 answers:
kaheart [24]2 years ago
6 0

Answer:

a = 2.22 [m/s^2]

Explanation:

First we have to convert from kilometers per hour to meters per second

40 [\frac{km}{h}]*[\frac{1h}{3600s}]*[\frac{1000m}{1km}] = 11.11 [m/s]

We have to use the following kinematics equation:

v_{f}= v_{i}+(a*t)

where:

Vf = final velocity = 11.11 [m/s]

Vi = initial velocity = 0

a = acceleration [m/s^2]

t = time = 5 [s]

The initial speed is taken as zero, as the car starts from zero.

11.11 = 0 + (a*5)

a = 2.22 [m/s^2]

Aliun [14]2 years ago
5 0

Refer to the attachment.

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The force needed to keep a car from skidding on a curve varies inversely as the radius of the curve and jointly as the weight of
motikmotik

Explanation:

It is given that, the force needed to keep a car from skidding on a curve varies inversely as the radius of the curve and jointly as the weight of the car and the square of the car's speed such that,

F\propto \dfrac{mgv^2}{r}

F=\dfrac{kmgv^2}{r}

mg is the weight of the car

r is the radius of the curve

v is the speed of the car

Case 1.

F = 640 pounds

Weight of the car, W = mg = 2600 pound

Radius of the curve, r = 650 ft

Speed of the car, v = 40 mph

640=\dfrac{k(2600)(40)^2}{650}

k = 0.1

Case 2.

Radius of the curve, r = 750 ft

Speed of the car, v = 30 mph

F=\dfrac{0.1\times 2600\times (30)^2}{750}

F = 312 N

Hence, this is the required solution.

6 0
3 years ago
When we experience positive "g forces", it is as if we have become...
zhenek [66]

Answer:

heavier

Explanation:

7 0
2 years ago
A strand of 10 lights is plugged into a outlet. How can you determine if the lights are connected in series or parrel
NeTakaya

Answer:

C) Unscrew one light. If the other lights turn off, it's a series circuit.

Explanation:

THIS IS THE COMPLETE QUESTION BELOW;

A strand of 10 lights is plugged into an outlet. How can you determine if the lights are connected in series or parallel? A) Unscrew one light. If the other lights stay on, it's a series circuit. B) Unplug the strand. If the first light stays on, it's a series circuit. C) Unscrew one light. If the other lights turn off, it's a series circuit. D) Cut the strand in half. If the plugged in half stays on, it's a series circuit.

SERIES CIRCUIT

In this circuit, the components there are in the same path, the entire circuit has the same current, each of the components posses different voltage drop. Hence, failure of one components to work, there will be break in entire circuit then other components cease to work.

PARALLEL CIRCUIT

This circuit has equal voltage drop across all the components, any problem in a component will not has effect on other components.

Therefore, if one want to determine if a light connection is in series or in parallel, one of the light can be unplugged if others stop working it means it's series, if other works it's parallel.

5 0
2 years ago
The orbital period of a satellite is 2 × 106 s and its total radius is 2.5 × 1012 m. The tangential speed of the satellite, writ
LenaWriter [7]

The orbital period of the satellite[T] is given as 2*10^{6} S.

The radius of the satellite is given [R] 2.5*10^{12} m.

we are asked here to calculate the tangential speed of the satellite.

Before going to get the solution first we have understand the tangential speed.

The tangential speed of a satellite is given as the speed required to keep the satellite along the orbit. If satellite speed is less than tangential speed,there is the chance of it falling down towards earth. If it is more,then it will deviate from it orbit and can't stick to the orbit further.In a simple way  the tangential speed is the linear speed of an object in a circular path.

Now we have to calculate the tangential speed [V].

Mathematically the tangential speed [V]   written as -

                                V=\frac{2\pi R}{T}

where T is the time period of the satellite and R is the radius of the satellite.

                        V=\frac{2*3.14*10^{12} }{2*10^{6} }

                               = 7.85*10^{6} m/s

There is also another way through which we can get  the solution as explained below-

We know that the tangential speed of a satellite V=\sqrt{\frac{GM}{R^{2} } }

where G is the gravitational constant and M is the mas of central object.

But we know that g=\frac{GM}{R^{2} }

                               ⇒GM=gR^{2}  where g is the acceleration due to gravity of that central object.


Hence    V=\sqrt{\frac{gR^{2} }{R} }

               ⇒   V=\sqrt{gR}

By knowing the value of g due to that central object we can also calculate its tangential speed.

                           

 




7 0
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