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3 years ago

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When I wave a charged golf tube at the front of the classroom with a frequency of two oscillations per second, I produce an elec

tromagnetic wave of frequency 2Hz. What is the wavelength of this wave? (a) 6.7 × 10−9m (b) 1.5 × 108m (c) 3.0 × 108m (d) 6.0 × 108m

1 answer:

solmaris [256]3 years ago

3 0

To solve this problem it is necessary to apply the concepts related to Wavelength depending on the frequency and speed of light.

The equation that meets these parameters is given by

$\lambda = \frac{c}{f}$

Where,

c = Speed of light

$\lambda = Wavelength$

f = Frequency

Our values are given as

$c = 3*10^8m/s \rightarrow$ velocity of light in free space

f = 2Hz

Replacing we have that

$\lambda = \frac{c}{f}\\\lambda = \frac{3*10^8}{2}\\\lambda = 1.5*10^8m$

Therefore the correct answer is B.

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For the ballistic missile aimed to achieve the maximum range of 9500 km, what is the maximum altitude reached in the trajectory

liraira [26]

Explanation:

The range <em>R</em> of a projectile is given the equation

$R = \dfrac{v_0^2}{g}\sin{2\theta}$

The maximum range is achieved when $\theta = 45°$ so our equation reduces to

$R_{max} = \dfrac{v_0^2}{g}$

We can solve for the initial velocity $v_0$ as follows:

$v_0^2 = gR_{max} \Rightarrow v_0 = \sqrt{gR_{max}}$

or

$v_0 = \sqrt{(9.8\:\text{m/s}^2)(9.5×10^6\:\text{m})}$

$\:\:\:\:\:\:\:=9.6×10^3\:\text{m/s}$

To find the maximum altitude H reached by the missile, we can use the equation

$v_y^2 = v_{0y}^2 - 2gy = (v_0\sin{45°})^2 - 2gy$

At its maximum height H, $v_y = 0$ so we can write

$0 = (v_0\sin{45°})^2 - 2gH$

or

$H = \dfrac{(v_0\sin{45°})^2}{2g}$

$\:\:\:\:\:\:= \dfrac{[(9.6×10^3\:\text{m/s})\sin{45°}]^2}{2(9.8\:\text{m/s}^2)}$

$\:\:\:\:\:\:= 2.4×10^6\:\text{m}$

7 0

2 years ago

A ball is tossed straight up from the surface of a small, sphericalasteroid with no atmosphere. The aball rises to a height equa

Anastaziya [24]

Answer:

1. 3) only a constant gravitation force that acts downward

2. b) Equal to the accleration at the surface of the asteroidc)

Explanation:

1.

In absence of atmosphere, the force that will act on the steroid is the force of gravity due to asteroid.

Gravitational force is a long range force acting always attractive in nature.

It is a contact-less field force which does not requires a medium.

Mathematically given as:

$F=G.\frac{M_1.M_2}{R^2}$

where:

G = gravitational constant

$M_1\ \&\ M_2$ are the mass of the two objects

$R=$ radial distance between the two objects.

2.

The acceleration of the ball at the top height of the path is still under the influence of the gravity of the two masses so it will be equal to the acceleration due to gravity at the surface of the asteroid. Acceleration always remains constant.

4 0

3 years ago

Rachel has been reading her physics book. She takes her weighing scales into an elevator and stands on them. If her normal weigh

GrogVix [38]

Answer:

345 N

Explanation:

Given:

Normal weight of Rachel (mg) = 690 N

Case 1: Upward motion of elevator

Given:

Acceleration of elevator (a) = 0.25 g

The scale reading is given by the normal force acting on Rachel. Let N₁ be the normal force.

So, the net force acting on Rachel is given as:

$F_{net}=N_1-mg=N_1-690$

Now, from Newton's second law:

$F_{net}=ma\\\\N_1-690=m\times 0.25g\\\\N_1-690=0.25\times (mg)\\\\N_1-690=0.25\times 690\\\\N_1=690+172.5=862.5\ N------(1)$

Case 2: Downward motion of elevator

Given:

Acceleration of elevator (a) = 0.25 g

The scale reading is given by the normal force acting on Rachel. Let N₂ be the normal force.

So, the net force acting on Rachel is given as:

$F_{net}=mg-N_2=690-N_2$

Now, from Newton's second law:

$F_{net}=ma\\\\690-N_2=m\times 0.25g\\\\690-N_2=0.25\times (mg)\\\\690-N_2=0.25\times 690\\\\N_2=690-172.5=517.5\ N------(2)$

Now, the difference in the scale reading is obtained by subtracting equation (2) from equation (1). This gives,

$Difference=N_1-N_2=862.5-517.5=345\ N$

Therefore, the difference between the up and down scale readings is 345 N.

4 0

3 years ago

A car is strapped to a rocket (combined mass = 661 kg), and its kinetic energy is 66,120 J.

Katen [24]

Answer:

9.4 m/s

Explanation:

The work-energy theorem states that the work done on an object is equal to the change in kinetic energy of the object.

So we can write:

$W=K_f - K_i$

where in this problem:

W = -36.733 J is the work performed on the car (negative because its direction is opposite to the motion of the car)

$K_i = 66,120 J$ is the initial kinetic energy of the car

$K_f$ is the final kinetic energy

Solving for Kf,

$K_f = W+K_i = -36,733+66,120=29,387 J$

The kinetic energy of the car can be also written as

$K_f = \frac{1}{2}mv^2$

where:

m = 661 kg is the mass of the car

v is its final speed

Solving, we find

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(29,387)}{661}}=9.4 m/s$

8 0

3 years ago

Charlie pulls horizontally to the right on a wagon with a force of 37.2 N. Sara pulls horizontally to the left with a force of 2

Sidana [21]

Answer:

The work done on the wagon is 37 joules.

Explanation:

Given that,

The force applied by Charlie to the right, F = 37.2 N

The force applied by Sara to the left, F' = 22.4 N

We need to find the work done on the wagon after it has moved 2.50 meters to the right. The net force acting on the wagon is :

$F_n=F-F'$

$F_n=37.2-22.4$

$F_n=14.8\ N$

Work done on the wagon is given by the product of net force and displacement. It is given by :

$W=F_n\times d$

$W=14.8\ N\times 2.5\ m$

W = 37 Joules

So, the work done on the wagon is 37 joules. Hence, this is the required solution.

7 0

3 years ago

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