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2 years ago

Select the correct answer.

Physics

1 answer:

Morgarella [4.7K]2 years ago

3 0

D all of them is the answer. All three can be used to celebrate sporting events

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The period of a sound wave is 0.002 seconds. The speed of a sound is 344 m/s. Find the frequency and wavelength of the sound wav

ankoles [38]

Answer:

I think that's the solution

5 0

3 years ago

The same wave travels for 2 seconds, what is its frequency?*

sveticcg [70]

Answer:

In waves distance is measured by wave length and time is measured by frequency or period.

velocity ratio=wave length multiply by frequency.

HENCE, if the same wave travels for 2 econds its frequency will be 2Hz.

Explanation:

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2 years ago

To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T

vladimir1956 [14]

Answer:

Explanation:

kinetic energy required = 1.80 MeV

= 1.8 x 10⁶ x 1.6 x 10⁻¹⁹ J

= 2.88 x 10⁻¹³ J

If v be the velocity of proton

1/2 x mass of proton x v² = 2.88 x 10⁻¹³

= .5 x 1.67 x 10⁻²⁷ x v² = 2.88 x 10⁻¹³

v² = 3.45 x 10¹⁴

v = 1.86 x 10⁷ m /s

If V be the potential difference required

V x e = kinetic energy . where e is charge on proton .

V x 1.6 x 10⁻¹⁹ = 2.88 x 10⁻¹³

V = 1.8 x 10⁶ volt .

3 0

3 years ago

1. A 14-cm tall object is placed 26 cm from a converging lens that has a focal length of 13 cm.

AURORKA [14]

Answer:

a) Please find attached the required drawing of light passing through the lens

By the use of similar triangles;

The image distance from the lens = 26 cm

The height of the image = 14 cm

c) The image distance from the lens = 26 cm

The height of the image = 14 cm

Explanation:

Question;

a) Determine the image distance and the height of the image

b) Calculate the image position and height

The given parameters are;

The height of the object, h = 14 cm

The distance of the object from the mirror, u = 26 cm

The focal length of the mirror, f = 13 cm

The location of the object = 2 × The focal length

Therefore, given that the center of curvature ≈ 2 × The focal length, we have;

The location of the object ≈ The center of curvature of the lens

The diagram of the object, lens and image created with MS Visio is attached

From the diagram, it can be observed, using similar triangles, that the image distance from the lens = The object distance from the lens = 26 m

The height of the image = The height of the object - 14 cm

b) The lens equation is used for finding the image distance from the lens as follows;

$\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}$

Where;

v = The image distance from the lens

We get;

$v = \dfrac{u \times f}{u - f}$

Therefore;

$v = \dfrac{26 \times 13}{26 - 13} = \dfrac{26 \times 13}{13} = 26$

The distance of the image from the lens, v = 26 cm

The magnification, M =v/u

∴ M = 26/26 = 1, therefore, the object and the image are the same size

Therefore;

The height of the image = The height of the object = 14 cm.

5 0

2 years ago

Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates

melisa1 [442]

Answer:

$v_{2.6b}=11.18\ m.s^{-1}$

$v_{7.2b}=14.19\ m.s^{-1}$

$s_{bb}=226.3305\ m$

$a_{db}=-3.7386\ m.s^{-2}$ negative sign denotes deceleration.

$t_b=21.3956\ s$

$a_y=1.1065\ m.s^{-2}$

Explanation:

Given:

initial speed of blue car, $u_b=0\ m.s^{-1}$

initial speed of yellow car, $u_y=0\ m.s^{-1}$

acceleration rate of blue car, $a_b=4.3\ m.s^{-2}$

time for which the blue car accelerates, $t_{ab}=3.3\ s$

time for which the blue car moves with uniform speed before decelerating, $t_{ub}=14.3\ s$

total distance covered by the blue car before coming to rest, $s_b=253.26 \ m$

distance at which the the yellow car intercepts the blue car just as the blue car come to rest, $s_y=253.26 \ m$

Speed of blue car after 2.6 seconds of starting the motion:

Applying the equation of motion:

$v_{2.6b}=u_b+a_b.t$

$v_{2.6b}=0+4.3\times 2.6$

$v_{2.6b}=11.18\ m.s^{-1}$

Speed of blue car after 7.2 seconds of starting the motion:

∵The car accelerates uniformly for 3.3 seconds after which its speed becomes uniform for the next 14.3 second before it applies the brake.

so,

$v_{7.2b}=u+a_b\times t_{ab}$

$v_{7.2b}=0+4.3\times 3.3$

$v_{7.2b}=14.19\ m.s^{-1}$

Distance travelled by the blue car before application of brakes:

This distance will be $s_{bb}=$ (distance travelled during the accelerated motion) + (distance travelled at uniform motion)

Now the distance travelled during the accelerated motion:

$s_{ab}=u_b.t_{ab}+\frac{1}{2} a_{b}.t_{ab}^2$

$s_{ab}=0\times 3.3+0.5\times 4.3\times 3.3^2$

$s_{ab}=23.4135\ m$

Now the distance travelled at uniform motion:

$s_{ub}=14.19\times 14.3$

$s_{ub}=202.917\ m$

Finally:

$s_{bb}=s_{ab}+s_{ub}$

$s_{bb}=23.4135+202.917$

$s_{bb}=226.3305\ m$

Acceleration of the blue car once the brakes are applied

Here we have:

initial velocity, $u=14.19\ m.s^{-1}$

final velocity, $v=0\ m.s^{-1}$

distance covered while deceleration, $s_{db}=s_b-s_{bb}$

$\Rightarrow s_{db}=253.26 -226.3305=26.9295\ m$

Using the equation of motion:

$v^2=u^2+2a_{db}.s_{db}$

$0^2=14.19^2+2\times a_{db}\times 26.9295$

$a_{db}=-3.7386\ m.s^{-2}$ negative sign denotes deceleration.

The total time for which the blue car moves:

$t_b=t_a+t_{ub}+t_{db}$ ........................(1)

Now the time taken to stop the blue car after application of brakes:

Using the eq. of motion:

$v=u+a_{db}.t_{db}$

$0=14.19-3.7386\times t_{db}$

$t_{db}=3.7956\ s$

Putting respective values in eq. (1)

$t_b=3.3+14.3+3.7956$

$t_b=21.3956\ s$

For the acceleration of the yellow car:

We apply the law of motion:

$s_y=u_y.t_y+\frac{1}{2} a_y.t_y^2$

Here the time taken by the yellow car is same for the same distance as it intercepts just before the stopping of blue car.

Now,

$253.26=0\times 21.3956+0.5\times a_y\times 21.3956^2$

$a_y=1.1065\ m.s^{-2}$

7 0

3 years ago