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3 years ago

How fast would 40 Newtons of force accelerate a 2 kg object?

Physics

1 answer:

Digiron [165]3 years ago

8 0

Answer:

$20 m/s^2$

Explanation:

We can solve this problem by using Newton's second law of motion, which states that the net force acting on an object is equal to the product between its mass and its acceleration:

$F=ma$

where

F is the net force on the object

m is its mass

a is its acceleration

In this problem:

F = 40 N is the force on the object

m = 2 kg is its mass

Therefore, the acceleration of the object is

$a=\frac{F}{m}=\frac{40}{2}=20 m/s^2$

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A speedboat increases its speed from 14.5 m/s to 29.3 m/s in a distance of 172 m.

dalvyx [7]

Answer:

a. Acceleration, a = 1.88 m/s²

b. Time, t = 7.87 seconds.

Explanation:

Given the following data;

Initial velocity, U = 14.5m/s

Final velocity, V = 29.3m/s

Distance, S = 172m

a. To find the acceleration of the speedboat;

We would use the third equation of motion;

V² = U² + 2aS

Substituting into the formula

29.3² = 14.5² + 2a*172

858.49 = 210.25 + 344a

344a = 858.49 - 210.25

344a = 648.24

a = 648.24/344

Acceleration, a = 1.88 m/s²

b. To find the time;

We would use the first equation of motion;

V = U + at

29.3 = 14.5 + 1.88t

1.88t = 29.3 - 14.5

1.88t = 14.8

Time, t = 14.8/1.88

Time, t = 7.87 seconds.

6 0

3 years ago

Two point charges are fixed on the y axis: a negative point charge q1 = -25 μC at y1 = +0.18 m and a positive point charge q2 at

dedylja [7]

Answer:

$50.91 \mu C$

Explanation:

The magnitude of the net force exerted on q is known, we have the values and positions for $q_{1}$ and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted by $q_{1}$ on q. Then we can know the magnitude of the force exerted by $q_{2}$ about q, finally this will allow us to know the magnitude of $q_{2}$

$q_{1}$ exerts a force on q in +y direction, and $q_{2}$ exerts a force on q in -y direction.

$F_{1}=\frac{kq_{1} q }{d^2}\\F_{1}=\frac{(8.99*10^9)(25*10^{-6}C)(8.4*10^{-6}C)}{(0.18m)^2}=58.26 N\\$

The net force on q is:

$F_{T}=F_{1} - F_{2}\\25N=58.26N-F_{2}\\F_{2}=58.26N-25N=33.26N\\\mid F_{2} \mid=\frac{kq_{2}q}{d^2}$

Rewriting for $q_{2}$ :

$q_{2}=\frac{F_{2}d^2}{kq}\\q_{2}=\frac{33.26N(0.34m)^2}{8.99*10^9\frac{Nm^2}{C^2}(8.4*10^{-6}C)}=50.91*10^{-6}C=50.91 \mu C$

8 0

3 years ago

Which electromagnetic wave has the lowest frequencies (less than 3×109 hertz)?

ICE Princess25 [194]

Answer:

Radio waves

Explanation:

The electromagnetic spectrum includes all different types of waves, which are usually classified depending on their frequency. Ordering them from the highest frequency to the lowest frequency, they are:

- Gamma rays

- X-rays

- Ultraviolet

- Visible light

- Infrared radiation

- Microwaves

- Radio waves

Radio waves are the electromagnetic waves with lowest frequency, their frequency is lower than 300 GHz ( $3\cdot 10^{11} Hz$ ) and therefore they are the electromagnetic waves with lowest energy (in fact, the energy of an electromagnetic wave is proportional to its frequency). They are generally used for radio and telecommunications since this type of waves can travel up to long distances.

5 0

4 years ago

Read 2 more answers

g An electron enters a region of space containing a uniform 1.63 × 10 − 5 T magnetic field. Its speed is 121 m/s and it enters p

kolbaska11 [484]

Answer:

i. The radius 'r' of the electron's path is 4.23 × $10^{-5}$ m.

ii. The frequency 'f' of the motion is 455.44 KHz.

Explanation:

The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.

r = $\frac{mv}{qB}$

Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.

From the question, B = 1.63 × $10^{-5}$ T, v = 121 m/s, Θ = $90^{0}$ (since it enters perpendicularly to the field), q = e = 1.6 × $10^{-19}$ C and m = 9.11 × $10^{-31}$ Kg.