Yo pls help me I'm desperate lowkey

9- Where did Rutherford propose that most of the mass of an atom

siniylev [52]

Answer:

A) in the middle of the atom

Explanation:

He concluded that all of the positive charge and the majority of the mass of the atom must be concentrated in a very small space in the atom's interior, which he called the nucleus.

8 0

3 years ago

g While hauling a log in the back of a flatbed truck, a driver is pulled over by the state police. Although the log cannot roll

irga5000 [103]

Answer:

The minimum coefficient of static friction required, µ = 0.10

Note. The question is incomplete. The complete question is given below:

While hauling a log in the back of a flatbed truck, a driver is pulled over by the state police. Although the log cannot roll sideways, the police claim that the log could have slid out the back of the truck when accelerating from rest. The driver claims that the truck could not possibly accelerate at the level needed to achieve such an effect. Regardless, the police write a ticket anyway and now the driver court date is approaching.

The log has a mass of m = 929 kg; the truck has a mass of M = 8850 kg. According to the truck manufacturer, the truck can accelerate from 0 to 55 mph in 23.0 seconds, but this does not account for the additional mass of the log. Calculate the minimum coefficient of static friction μs needed to keep the log in the back of the truck.

Explanation:

First, velocity in mph is converted to m/s

1 mph = 0.447 m/s

55 mph ≈ 24.6 m/s

The acceleration of the empty truck is a = v/t = 24.6 / 23 = 1.07 m/s²

Force that can be generated by the truck, F = ma

F = 8850kg * 1.07 m/s² = 9469.5 N

However, with the added mass of the log on it, the acceleration of the truck will become;

a = F / m = 9469.5 N / (8550+929)kg = 0.97 m/s²

Frictional force between the log and the truck = 0.97 m/s² * 929 kg = 901.13 N

Normal reaction on the truck due to the weight of the log, R = mg

R = 929 kg * 9.8m/s² = 9104.2 N

Coefficient of static friction, µ = F/R

µ = 901.13/9104.2

µ = 0.098 ≈ 0.10

Therefore, the minimum static friction required is µ = 0.10

8 0

3 years ago

Which has more kinetic energy, a 40 kg cheetah running at 25 m/s or a 4,000 kg elephant moving at 2 m/s? How much more energy do

denis23 [38]

Answer:

1st case; mass=40 kg

v=25m/s

K.E.=1/2[40][25]^2=12,500 J

2nd case; mass=4,000 kg

v=2 m/s

K.E.=1/2[4000][2]^2=8000J

in case 1 Kinetic energy is greater.

Explanation:

8 0

3 years ago

1

marishachu [46]

Answer: B

Explanation:

4 0

3 years ago

A skateboarder travels on a horizontal surface with an initial velocity of 3.6 m/s toward the south and a constant acceleration

Dimas [21]

Answer:

a) The x-position of the skateboarder is 0.324 meters.

b) The y-position of the skateboarder is -2.16 meters.

c) The x-velocity of the skateboard is 1.08 meters per second.

d) The y-velocity of the skateboard is -3.6 meters per second.

Explanation:

a) The x-position of the skateboarder is determined by the following expression:

$x(t) = x_{o} + v_{o,x}\cdot t + \frac{1}{2}\cdot a_{x} \cdot t^{2}$ (1)

Where:

$x_{o}$ - Initial x-position, in meters.

$v_{o,x}$ - Initial x-velocity, in meters per second.

$t$ - Time, in seconds.

$a_{x}$ - x-acceleration, in meters per second.

If we know that $x_{o} = 0\,m$ , $v_{o,x} = 0\,\frac{m}{s}$ , $t = 0.60\,s$ and $a_{x} = 1.8\,\frac{m}{s^{2}}$ , then the x-position of the skateboarder is:

$x(t) = 0\,m + \left(0\,\frac{m}{s} \right)\cdot (0.60\,s) + \frac{1}{2}\cdot \left(1.8\,\frac{m}{s^{2}} \right) \cdot (0.60\,s)^{2}$

$x(t) = 0.324\,m$

The x-position of the skateboarder is 0.324 meters.

b) The y-position of the skateboarder is determined by the following expression:

$y(t) = y_{o} + v_{o,y}\cdot t + \frac{1}{2}\cdot a_{y} \cdot t^{2}$ (2)

Where:

$y_{o}$ - Initial y-position, in meters.

$v_{o,y}$ - Initial y-velocity, in meters per second.

$t$ - Time, in seconds.

$a_{y}$ - y-acceleration, in meters per second.

If we know that $y_{o} = 0\,m$ , $v_{o,y} = -3.6\,\frac{m}{s}$ , $t = 0.60\,s$ and $a_{y} = 0\,\frac{m}{s^{2}}$ , then the x-position of the skateboarder is:

$y(t) = 0\,m + \left(-3.6\,\frac{m}{s} \right)\cdot (0.60\,s) + \frac{1}{2}\cdot \left(0\,\frac{m}{s^{2}}\right)\cdot (0.60\,s)^{2}$

$y(t) = -2.16\,m$

The y-position of the skateboarder is -2.16 meters.

c) The x-velocity of the skateboarder ( $v_{x}$ ), in meters per second, is calculated by this kinematic formula:

$v_{x}(t) = v_{o,x} + a_{x}\cdot t$ (3)

If we know that $v_{o,x} = 0\,\frac{m}{s}$ , $t = 0.60\,s$ and $a_{x} = 1.8\,\frac{m}{s^{2}}$ , then the x-velocity of the skateboarder is:

$v_{x}(t) = \left(0\,\frac{m}{s} \right) + \left(1.8\,\frac{m}{s} \right)\cdot (0.60\,s)$

$v_{x}(t) = 1.08\,\frac{m}{s}$

The x-velocity of the skateboard is 1.08 meters per second.

d) As the skateboarder has a constant y-velocity, then we have the following answer:

$v_{y} = -3.6\,\frac{m}{s}$

The y-velocity of the skateboard is -3.6 meters per second.

3 0

3 years ago