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4 years ago

What are atoms composed of

Physics

2 answers:

navik [9.2K]4 years ago

8 0

Answer: Atoms are composed of protons, electrons and neutrons.

Explanation:

madam [21]4 years ago

7 0

Protons,electrons, and neutrons

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The velocity v of a particle moving in the xy plane is given by v = (6.0t -4.0t2 )i + 8.5j, in m/s. Here v is in m/s and t (for

atroni [7]

Answer:

Explanation:

Given

Velocity of the particle in vector form is given by

$v=\left ( 6t-4t^2\right )\hat{i}+8.5\hat{j}$

acceleration is rate of change of velocity thus acceleration is

$a=\frac{\mathrm{d} v}{\mathrm{d} t}$

$a=(6-8t)\hat{i}+0\hat{j}$

at $t=3\ s$

$a=6-8\times 3$

$a=-18\ m/s^2$

7 0

4 years ago

What is the speed of a transverse wave in a rope of length 2.00 m and mass 60.0 g under a tension of 500 n?

drek231 [11]

The formula we can use in this case would be:

v = sqrt (T / (m / l))

Where,

v = is the velocity of the transverse wave = unknown (?)

T = is the tension on the rope = 500 N

m = is the mass of the rope = 60.0 g = 0.06 kg

l = is the length of the rope = 2.00 m

Substituting the given values into the equation to search for the speed v:
v = sqrt (500 N/(0.06 kg /2 m))
v = sqrt (500 * 2 / 0.06)
v = sqrt (16,666.67)
<span>v = 129.10 m/s</span>

7 0

3 years ago

A 5.0 g coin is placed 15 cm from the center of a turntable. The coin has static and kinetic coefficients of friction with the t

Kaylis [27]

Answer:

Angular speed will reach 6.833rad/s before the coin starts slipping

Explanation:

There is no question but I'll asume the common one: Calculate the speed of the turntable before the coin starts slipping.

With a sum of forces:

$Ff = m*a$

$Ff=m*V^2/R$

At this point, friction force is maximum, so:

$\mu*N=m*V^2/R$

$\mu*m*g=m*V^2/R$

Solving for V:

$V=\sqrt{\mu*g*R}$

V=1.025 m/s

The angular speed of the turntable will be:

ω = V/R = 6.833 rad/s This is the maximum speed it can reach before the coin starts slipping.

7 0

3 years ago

A car of 1000 kg with good tires on a dry road can decelerate (slow down) at a steady rate of about 5.0 m/s2 when braking. If a

vredina [299]

(a) 4.0 s

The acceleration of the car is given by

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time interval

For this car, we have

v = 0 (the final speed is zero since the car comes to a stop)

u = 20 m/s is the initial velocity

$a=-5.0 m/s^2$ is the deceleration of the car

Solving the equation for t, we find the time needed to stop the car:

$t=\frac{v-u}{a}=\frac{0-(20 m/s)}{-5.0 m/s^2}=4 s$

(b) 40 m

The stopping distance of the car can be calculated by using the equation

$v^2 - u^2 = 2ad$

where

v = 0 is the final velocity

u = 20 m/s is the initial velocity

a = -5.0 m/s^2 is the acceleration of the car

d is the stopping distance

Solving the equation for d, we find

$d=\frac{v^2-u^2}{2a}=\frac{0^2-(20 m/s)^2}{2(-5.0 m/s^2)}=40 m$

(c) $-5.0 m/s^2$

The deceleration is given by the problem, and its value is $-5.0 m/s^2$ .

(d) 5000 N

The net force applied on the car is given by

$F=ma$

where

m is the mass of the car

a is the magnitude of the acceleration

For this car, we have

m = 1000 kg is the mass

$a=5.0 m/s^2$ is the magnitude of the acceleration

Solving the formula, we find

$F=(1000 kg)(5.0 m/s^2)=5000 N$

(e) $2.0\cdot 10^5 J$

The work done by the force applied by the car is

$W=Fd$

where

F is the force applied

d is the total distance covered

Here we have

F = 5000 N

d = 40 m (stopping distance)

So, the work done is

$W=(5000 N)(40 m)=2.0\cdot 10^5 J$

(f) The kinetic energy is converted into thermal energy

Explanation:

when the breaks are applied, the wheels stop rotating. The car slows down, as a result of the frictional forces between the brakes and the tires and between the tires and the road. Due to the presence of these frictional forces, the kinetic energy is converted into thermal energy/heat, until the kinetic energy of the car becomes zero (this occurs when the car comes to a stop, when v = 0).

8 0

4 years ago

A force of 120 N is exerted on a 40 kg container which sits on a floor. If the frictional force between floor and container is 8

harkovskaia [24]

Answer:

Magnitude is 144N .

Explanation:

a²+b²=c²

120N²+80N²=c²

14400+6400=c²

$\sqrt{20800} = \sqrt{c} ^{2}$

144N=c

7 0

3 years ago