All categories

2 years ago

A race car has a maximum speed of 0.104 km/s .What is this speed in miles per hour ?

Physics

1 answer:

sweet [91]2 years ago

8 0

Answer:

232.641374 mph

Explanation:

A race car has a maximum speed of 0.104km/s

Let X represent the speed in miles per hour

Therefore the speed in miles per hour can be calculated as follows

1 km/s = 2,236.936292 mph

0.104km/s = X

X = 0.104 × 2,236.936292

X = 232.641374

Hence the speed in miles per hour is 232.641374 mph

You might be interested in

In this lab you will use a cart and a track to explore Newton's second law of motion. You will vary two different variables and

Savatey [412]

Answer:

Newtons II law:

It is defined as "the net force acting on the object is a product of mass and acceleration of the body" . Also it defines that the "acceleration of an object is dependent on net force and mass of the body".

Let us assume that,a string is attached to the cart, which passes over a pulley along the track. At another end of the string a weight is attached which hangs over the pulley. The hanging weight provides tension in the spring, and it helps in accelerating the cart. We assume that the string is massless and no friction between pulley and the string.

Whenever the hanging weight moves downwards, the cart will accelerate to right side.

For the hanging weight/mass

When hanging weight of mass is m₁ and accelerate due to gravitational force g.

Therefore we can write F = m₁ .g

and the tension acts in upward direction T (negetive)

Now, Fnet = m₁ .g - T

= m₁.a

So From Newtons II law F = m.a

3 0

2 years ago

Read 2 more answers

A uniformly accelerated car passes three equally spaced traffic signs. The signs are separated by a distance d = 25 m. The car p

DedPeter [7]

Answer:

a) $v_{1}=\frac{x_{2}-x_{1} }{t_{2}-t_{1} }=\frac{(2(\frac{25}{3})-\frac{25}{3} )m}{3.9s-1.3s} =3.2051 \frac{m}{s}$

b) $v_{2}=\frac{x_{3}-x_{2} }{t_{3}-t_{2} }=\frac{(25-2(\frac{25}{3}) )m}{5.5s-3.9s} =5.2083 \frac{m}{s}$

c) $a=\frac{v_{2}-v_{1} }{t_{2}-t_{1} } =\frac{5.2083m/s-3.2051m/s}{5.5s-3.9s} =1.252 \frac{m}{s^{2} }$

Explanation:

The knowable variables are

$d_{t}=25m$

$t_{1}=1.3 s$

$t_{2}=3.9 s$

$t_{3}=5.5 s$

Since the three traffic signs are equally spaced, the distance between each sign is $\frac{25}{3} m$

a) $v_{1}=\frac{x_{2}-x_{1} }{t_{2}-t_{1} }=\frac{(2(\frac{25}{3})-\frac{25}{3} )m}{3.9s-1.3s} =3.2051 \frac{m}{s}$

b) $v_{2}=\frac{x_{3}-x_{2} }{t_{3}-t_{2} }=\frac{(25-2(\frac{25}{3}) )m}{5.5s-3.9s} =5.2083 \frac{m}{s}$

Since we know the velocity in two points and the time the car takes to pass the traffic signs

c) $a=\frac{v_{2}-v_{1} }{t_{2}-t_{1} } =\frac{5.2083m/s-3.2051m/s}{5.5s-3.9s} =1.252 \frac{m}{s^{2} }$

6 0

2 years ago

If an electromagnetic wave has components Ey = E0 sin(kx - ωt) and Bz = B0 sin(kx - ωt), in what direction is it traveling?

fomenos

Answer:

Its traveling in the +x direction

Explanation:

The E-field is in the +y-direction, and the B-field is in the +z-direction, so it must be moving along the +x-direction, since the E-field, B-field and the direction of moving are all at right angles to each other.

8 0

2 years ago

A seasoned mini golfer is trying to make par on a tricky number five hole. The golfer can complete the hole by hitting the ball

liberstina [14]

Answer:5.17 m/s

Explanation:

Given

let u be the speed at cliff initial point

range over cliff is 1.45 m

and range of projectile is given by

$R=\frac{u^2\sin 2\theta }{g}$