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3 years ago

What force causes a 1-kg mass to accelerate at a rate of 1 meter per second each second?

Physics

1 answer:

Zigmanuir [339]3 years ago

8 0

Answer:

$F=1N$

Explanation:

Conceptual analysis

To solve this problem we apply Newton's second law:

The acceleration of an object is proportional to the force F acting on it and inversely proportional to its mass m.

a = F / m

Where,

F = m * a Formula (1)

F: Force in Newtons (N)

m: mass in kg

a: acceleration in m/(s^2)

a = v / t Formula (2)

v: speed in m/s

t: time in seconds (s)

Known information

We know the following data:

m = 1kg

v = 1 m/s

t = 1s

Development of the problem:

In the Formula (2): $a = \frac{\frac{1m}{s}}{1s} = 1 \frac{m}{s^2}$

In the Formula (1): $F=1kg* 1 \frac{m}{s^2}=1N$

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Two boxes are connected to each other as shown. The system is released from rest and the 1.00-kg box falls through a distance of

disa [49]

Answer:

The kinetic energy of box b is 2.44 J.

Explanation:

Given that,

Mass of box a =3.00 kg

Mass of box b=1.00 kg

Distance = 1.00 m

We need to calculate the acceleration of the system

Using formula of acceleration

$a=\dfrac{m_{2}\times g}{m_{1}+m_{2}}$

put the value into the formula

$a=\dfrac{1.00\times9.8}{1.00+3.00}$

$a=2.45\ m/s^2$

We need to calculate the velocity

Using equation of motion

$v^2=u^2+2as$

Put the value into the formula

$v^2=0+2\times2.45\times1.00$

$v=\sqrt{2\times2.45\times1.00}$

$v=2.21\ m/s$

We need to calculate the kinetic energy of box b

Using formula of energy

$K.E=\dfrac{1}{2}mv^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times1.00\times(2.21)^2$

$K.E=2.44\ J$

Hence, The kinetic energy of box b is 2.44 J.

3 0

4 years ago

Newton discovered that sunlight was composed of colors.<br><br><br> True or False

KonstantinChe [14]

Answer:

True

Newton's Rainbow. In the 1660s, English physicist and mathematician Isaac Newton began a series of experiments with sunlight and prisms. He demonstrated that clear white light was composed of seven visible colors.

Explanation:

Hope it helps

5 0

3 years ago

A 0.300 kg potato is tied to a string with length 2.30 m , and the other end of the string is tied to a rigid support. The potat

alexdok [17]

Answer:

The speed of the potato at the lowest point is 6.714 m/s

The tension in the string is 8.82 N

Explanation:

We will use law of conservation of energy to find out the velocity

When the potato is at 2.30 m height, it will have potential energy and when the potato released it will go lowest point and it's potential energy will be convert into kinetic energy.

Potential energy = Kinetic energy

m×g×h = (1/2) m×v²

m stands for mass

g stands for gravitational constant

h stands for height

v stands for velocity

m×g×h = (1/2) m×v²

0.3×9.8×2.3 = (1/2) .3×v²

v² = 45.08

v = 6.714 m/s

The speed of the potato at lowest point is 6.714 m/s

Part B

The potato is in uniform circular motion so the acceleration is given by

a = v² / r

a stands for centripetal acceleration

According to Newton second law the acceleration is directly proportional to applied force and inversely proportional to the mass of object

So ∑ F = m × a

Two forces act on potato 1. weight due to gravitation in downward direction 2. tension in upward direction So the above equation become

Tension force - Force due to weight = m×a

T-m×g = m×a

centripetal acceleration formula =v²/r , r = l = length of the string

T = m ×(v²/l) + m×g

v = 6.714 m/s , l = 2.30 m , m = .3 kg Now put values in above equation

T = .3 × (6.714²/2.30) + .3 × 9.8

T = .3 ( 45.08/2.30) + 2.94

T = .3 (19.6) +2.94

T = 5.88 + 2.94

T = 8.82 N

The tension in the string is 8.82 N

5 0

3 years ago

Four spheres with positive and negative charges hang from strings.

Arada [10]

Be because it’s the answer

5 0

3 years ago

Monochromatic light falls on a slit that is 2.30×10−3mm wide. Part A If the angle between the first dark fringes on either side

slava [35]

Answer:

Wavelength, $\lambda=1.28\times 10^{-6}\ m$

Explanation:

Given that,

Width of the slit, $d=2.3\times 10^{-3}\ mm=2.3\times 10^{-6}\ m$

The angle between the first dark fringes on either side of the central maximum is 34 degrees

To find,

The wavelength of the light used.

Solution,

The equation of maxima is given by :

$dsin\theta=n\lambda$

n = 1 here

$\lambda=\dfrac{d\ sin\theta}{n}$

$\lambda={d\ sin\theta}$

$\lambda={2.3\times 10^{-6}\times sin(34)}$

$\lambda=1.28\times 10^{-6}\ m$

So, the wavelength of the light used is $1.28\times 10^{-6}\ m$ . Hence, this is the required solution.

5 0

3 years ago