If you are traveling at 75 km/h how long will it take to travel 32 km?

Calculate the size of the image of a tree that is 8m high and 80 m a pinhole camera that is 20 cm long . what is its magnificati

Vladimir79 [104]

1) Size of the image: 2 cm

In order to calculate the size of the image, we can use the following proportion:

$p:q = h_o : h_i$

where

p = 80 m is the distance of the tree from the pinhole

q = 20 cm = 0.2 m is the distance of the image from the pinhole

$h_o$ = 8 m is the heigth of the object

$h_i$ is the height of the image

By re-arranging the proportion, we find

$h_i = \frac{h_o \cdot q}{p}=\frac{(8 m)(0.2 m)}{80 m}=0.02 m=2 cm$

2) Magnification: 0.0025

The magnification of a camera is given by the ratio between the size of the image and the size of the real object:

$M=\frac{h_i}{h_o}$

so, in this problem we have

$M=\frac{0.02 m}{8 m}=0.0025$

4 0

3 years ago

To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following general equation y(

DerKrebs [107]

Answer:

0.0549 m

Explanation:

Given that

equation y(x,t)=Acos(kx−ωt)

speed v = 8.5 m/s

amplitude A = 5.5*10^−2 m

wavelength λ = 0.5 m

transverse displacement = ?

v = angular frequency / wave number

and

wave number = 2π/ λ

wave number = 2 * 3.142 / 0.5

wave number = 12.568

angular frequency = v k

angular frequency = 8.5 * 12.568

angular frequency = 106.828 rad/sec ~= 107 rad/sec

so

equation y(x,t)=Acos(kx−ωt)

y(x,t)= 5.5*10^−2 cos(12.568 x−107t)

when x =0 and and t = 0

maximum y(x,t)= 5.5*10^−2 cos(12.568 (0) − 107 (0))

maximum y(x,t)= 5.5*10^−2 m

and when x = x = 1.52 m and t = 0.150 s

y(x,t)= 5.5*10^−2 cos(12.568 (1.52) −107(0.150) )

y(x,t)= 5.5*10^−2 × (0.9986)

y(x,t) = 0.0549 m

so the transverse displacement is 0.0549 m

5 0

3 years ago

Two objects with masses m and 3m are connected by a compressed spring so that the energy stored in the spring is 375 joules. If

cluponka [151]

Answer:

281.25 J

Explanation:

We are told that the two objects with masses m and 3m.

Also that energy stored in the spring is 375 joules.

Now, initially the centre of mass of the system took place at rest, it means v1 = v and v2 = v/3

Thus, from principle of conservation of energy, we have;

½mv² + ½(3m)(v/3)² = 375J

(m + 3m/9)½v² = 375

(4/3)m × ½v² = 375

Multiply both sides by ¾ to get;

½mv² = 375 × ¾

½mv² = 281.25 J

Therefore, energy of lighter body is 281.25 J

7 0

2 years ago

1) A pendulum is configured to have a period of 2 seconds.

vazorg [7]

"The" (and any subsequent words) was ignored because we limit queries to 32 words.

3 0

3 years ago

A spring with constant k = 78 N/m is at the base of a frictionless, 30.0°-inclined plane. A 0.50-kg block is pressed against the

olasank [31]

Explanation:

The given data is as follows.

Spring constant (k) = 78 N/m, $\theta = 30^{o}$

Mass of block (m) = 0.50 kg

According to the formula of energy conservation,

mgh sin $\theta = \frac{1}{2}kx^{2}$

h = $\frac{1}{2} \times \frac{kx^{2}}{mg Sin \theta}$

= $\frac{78 \times 0.04}{2 \times 0.5 \times 9.8 \times 0.5}$

= 0.64 m

Thus, we can conclude that the distance traveled by the block is 0.64 m.

4 0

3 years ago