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3 years ago

How much force is needed to accelerate a 20 kg mass at a rate of 4 m/s to the second power?

Physics

2 answers:

LekaFEV [45]3 years ago

8 0

Answer:

80 N

Explanation:

mass (m) = 20kg

acceleration (a) = 4m/s^2

Force (F) = ?

.°. F = ma

= 20 × 4 = 80

.°. F = 80 N

Thus, The Force is needed to accelerate a 20 kg mass at a rate of 4 m/s^2 is 80 N

-TheUnknownScientist

dusya [7]3 years ago

6 0

Answer:

So 55 Newtons are needed.

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A radioactive decay is illustrated. Which radioactive emission is a product of this decay?

hram777 [196]

The radioactive decay is shown in the image attached to this answer

Answer:

Alpha

Explanation:

An alpha decay is said to have occurred when the mass number of the daughter nucleus decreases by four units and the atomic number of the daughter nucleus decreases by two units.

If we look at the equation from left to right, we will notice that the mass number of the daughter nucleus decreased by four units and the atomic of the daughter nucleus number by two units compared to that of the parent nucleus leading to the emission of an alpha particle.

5 0

3 years ago

1.60 kg frictionless block is attached to an ideal spring with force constant 315 N/m . Initially the spring is neither stretche

Tatiana [17]

Answer:

(a). The amplitude of the motion is 0.926 m.

(b). The block’s maximum acceleration is 182.31 m/s².

(c). The maximum force the spring exerts on the block is 291.69 N.

Explanation:

Given that,

Mass of block =1.60 kg

Force constant = 315 N/m

Speed = 13.0 m/s

(a). We need to calculate the amplitude of the motion

Using conservation of energy

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2$

$A^2=\dfrac{mv^2}{k}$

Put the value into the relation

$A^2=\dfrac{1.60\times13.0^2}{315}$

$A=\sqrt{0.858}$

$A=0.926\ m$

(b). We need to calculate the block’s maximum acceleration

Using formula of acceleration

$a=A\omega^2$

$a=A\times\dfrac{k}{m}$

Put the value into the formula

$a=0.926\times\dfrac{315}{1.60}$

$a=182.31\ m/s^2$

(c). We need to calculate the maximum force the spring exerts on the block

Using formula of force

$F=ma$

Put the value into the formula

$F= 1.60\times182.31$

$F=291.69\ N$

Hence, (a). The amplitude of the motion is 0.926 m.

(b). The block’s maximum acceleration is 182.31 m/s².

(c). The maximum force the spring exerts on the block is 291.69 N.

7 0

3 years ago

"a horn emits a frequency of 1000 HZ. A 14 m/s wind is blowing toward a listener. What is the frequency of the sound heard by th

Semenov [28]

Answer:

$f_L = 1000 Hz$

Explanation:

given,

speed of wind, = 14 m/s

frequency of horn,f_s = 1000 Hz

speed of sound,V = 344 m/s

frequency heard by the listener

using Doppler effect

$f_L = \dfrac{v+v_L}{v+v_s}f_s$

f_L is the frequency of the sound heard by the listener

f_s is the frequency of sound emitted by the listener

V is the speed of sound

v_L is the speed of listener

v_s is the speed of source

now,

considering the frame of reference in which wind is at rest now, both listener and the source will be moving at 14 m/s

$f_L = \dfrac{v+14}{v+14}\times 1000$

now on solving we will get

$f_L = 1000 Hz$

hence, the frequency heard by the listener is equal to 1000 Hz

7 0

3 years ago

The area of the pond is approximately equal to the area of a circle with radius 297m. Find the mass of the ice. Answer in kilogr

True [87]

Answer:

mass of the ice is 254980463.8T kg

where T is the value of the thickness omitted in the question.

Explanation:

The ice on Walden Pond is .......... thick. The area of the pond is approximately equal to the area of a circle with radius 297 m. Find the mass of the ice. Answer in kg.

The value of the thickness of the ice T is omitted, but I will show the solution, and the real answer can be gotten by multiplying the final calculated answer here by the thickness of the ice omitted.

Given the radius of the equivalent circle of the ice = 297 m'

the area of the ice can be gotten from area A = $\pi r^{2}$ = $3.142*297^{2}$ = 277152.678 m^2

recall that the density of ice p ≅ 920 kg/m^3

also,

density of ice p = (mass of ice, m) ÷ (volume of ice, v)

i.e p = m/v

and,

m = pv

substituting the value of the density of water p into the equation, we have,

mass of the ice, m = 920v ....... equ 1

The volume of the ice above will be = (area of the ice, A) x (thickness of the ice, T)

i.e v = AT

substituting the value of area A into the equation, we have

v = 277152.678T ......equ 2

substitute value of v into equ 1

mass of the ice, m = 920 x (277152.678T)

mass of the ice, m = 254980463.8T kg

where T is the thickness of the ice

NB: To get the mass, multiply this answer with the thickness T given in the question.

7 0

3 years ago

A 0.50-kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.9 m on a frictionless horizontal su

Sedaia [141]

Answer:

$\boxed {\boxed {\sf 18 \ m/s}}$

Explanation:

The ball is moving in a circle, so the force is centripetal.

One formula for calculating centripetal force is:

$F_c= \frac{mv^2}r}$

The mass of the ball is 0.5 kilograms. The radius is 1.9 meters. The centripetal force is 85 Newtons or 85 kg*m/s².

$F_c$ = 85 kg*m/s²
m= 0.5 kg
r= 1.9 m

Substitute the values into the formula.

$85 \ kg*m/s^2 = \frac{0.5 \ kg *v^2}{1.9 \ m}$

Isolate the variable v. First, multiply both sides by 1.9 meters.

$(1.9 \ m)(85 \ kg*m/s^2) = \frac{0.5 \ kg *v^2}{1.9 \ m}*1.9 \ m$

$(1.9 \ m)(85 \ kg*m/s^2) = {0.5 \ kg *v^2}$

$161.5 \ kg*m^2/s^2 = 0.5 \ kg*v^2$

Divide both sides by 0.5 kilograms.

$\frac {161.5 \ kg*m^2/s^2}{0.5 \ kg} = \frac{0.5 \ kg*v^2}{0.5 \ kg}$

$\frac {161.5 \ kg*m^2/s^2}{0.5 \ kg} =v^2$

$323 \ m^2/s^2 = v^2$

Take the square root of both sides of the equation.

$\sqrt {323 \ m^2/s^2} =\sqrt{ v^2$

$\sqrt {323 \ m^2/s^2} =v$

$17.9722007556 \ m/s =v$

The original measurements have 2 significant figures, so our answer must have the same.

For the number we found, 2 sig fig is the ones place. The 9 in the tenth place tells us to round the 7 to an 8.

$18 \ m/s =v$

The maximum speed is approximately 18 meters per second.

8 0

3 years ago