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Mama L [17]
3 years ago
7

A certain light truck can go around a flat curve having a radius of 150 m with a maximum spee dof 32.0 m/s. With what maximum sp

eed can it go around a curve having a radius of 75.0 m?
Physics
1 answer:
Dvinal [7]3 years ago
7 0

Answer

Maximum speed at 75 m radius will be 22.625 m /sec

Explanation:              

We have given radius of the curve r = 150 m

Maximum speed v_{max}=32m/sec

Coefficient of friction \mu =\frac{v_{max}^2}{rg}=\frac{32^2}{150\times 9.8}=0.6965

Now new radius r = 75 m

So maximum speed at new radius v_{max}=\sqrt{\mu rg}=\sqrt{0.6965\times 75\times 9.8}=22.625m/sec

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Mathphys Help help help
Keith_Richards [23]

Answer:

1030 mph

Explanation:

The new velocity equals the initial velocity plus the wind velocity.

First, in the x (east) direction:

vₓ = 335 mph + 711 cos 19° mph

vₓ = 1007 mph

And in the y (north) direction:

vᵧ = 0 mph + 711 sin 19° mph

vᵧ = 231 mph

The net speed can be found with Pythagorean theorem:

v² = vₓ² + vᵧ²

v² = (1007 mph)² + (231 mph)²

v ≈ 1030 mph

6 0
3 years ago
During a supernova, the outer layers of a star are blown off and the star's core shrinks down by a factor of 10,000 or more to f
Lana71 [14]

Answer:

a. the core will spin faster.

Explanation:

By law of conservation of angular momentum

(mvR)i= (mvR)f

m= mass of star

v= speed of star

R= radius of star

i= initial

f= final

since, size(R) of the star is reduced by factor of 10,000 and mass remains the same, the velocity must increase by the same factor to keep the angular momentum conserved.

Hence, a. the core will spin faster.

8 0
3 years ago
The plate near the Artic Ridge is among those with the slower growth rates, moving slightly less than 2.5 cm/year. How much will
Aleks04 [339]
Greater than 25 . 2.5 x 10 = 30
4 0
3 years ago
Read 2 more answers
A camera with a 50.0-mm focal length lens is being used to photograph a person standing 3.00 m away. (a) How far from the lens m
kirill [66]

a) 50.8 mm

b) The whole image (1:1)

c) It seems reasonable

Explanation:

a)

To project the image on the film, the distance of the film from the lens must be equal to the distance of the image from the lens. This can be found by using the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length of the lens

p is the distance of the object from the lens

q is the distance of the image from the lens

In this problem:

f = 50.0 mm = 0.050 m is the focal length (positive for a convex lens)

p = 3.00 m is the distance of the person from the lens

Therefore, we can find q:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{0.050}-\frac{1}{3.00}=19.667m^{-1}\\q=\frac{1}{19.667}=0.051 m=50.8 mm

b)

Here we need to find the height of the image first.

This can be done by using the magnification equation:

\frac{y'}{y}=-\frac{q}{p}

where:

y' is the height of the image

y = 1.75 m is the height of the real person

q = 50.8 mm = 0.0508 m is the distance of the image from the lens

p = 3.00 m is the distance of the person from the lens

Solving for y', we find:

y'=-\frac{qy}{p}=-\frac{(0.0508)(1.75)}{3.00}=-0.0296 m=-29.6mm

(the negative sign means the image is inverted)

Therefore, the size of the image (29.6 mm) is smaller than the size of the film (36.0 mm), so the whole image can fit into the film.

c)

This seems reasonable: in fact, with a 50.0 mm focal length, if we try to take the picture of a person at a distance of 3.00 m, we are able to capture the whole image of the person in the photo.

3 0
3 years ago
If you know the distance traveled and the time traveled, you can determine an object’s
lara31 [8.8K]
If I tell you that I traveled 30 miles in the last 2 hours, the only thing you can calculate is my average speed during those two hours. You can't tell anything about my acceleration or the direction in which I traveled.
3 0
3 years ago
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