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2 years ago

Grant sprints 50m to the right with an average velocity of 3.0 m/s

Physics

1 answer:

zlopas [31]2 years ago

6 0

I need help with my math

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Two 1.0 kg masses are 4.0 m apart on a frictionless table. Each has 1.0μC of charge. Part A What is the magnitude of the electri

9966 [12]

Force between two charges is given by

$F =\frac{kq_1q_2}{r^2}$

$F =\frac{9*10^9* 1* 10^{-6}* 1 * 10^{-6}}{4^2}$

$F = 5.625 * 10^{-4} N$

Now in order to find the acceleration of each mass

we can use

F = ma

$5.625 * 10^{-4} = 1 * a$

$a= 5.625 * 10^{-4} m/s^2$

8 0

3 years ago

Read 2 more answers

Which of the following is/are ALWAYS true concerning collision and transition state theory?

Scrat [10]

Answer:

1) Transition states are short-lived

Explanation:

Transition state theory explains the rates of elementary chemical reactions. It assumes a quasi-equilibrium between reactants and activated transition state complexes.

The following are the characteristics of transition states

Instability
Ill-defined
High energy
short-lived

The species that must collide for the reaction to occur are shown by the mechanism of reaction and not the balanced reaction itself

Intermediates are consumed in each step of the overall reaction, they are not short lived

6 0

2 years ago

Please help me with this 29 points

Irina18 [472]

Answer:

)Give the definition of poverty line as defined by the World Bank.

7 0

2 years ago

If the coefficient of friction is 0.3900 and the cylinder has a radius of 2.700 m, what is the minimum angular speed of the cyli

Aleks04 [339]

Answer:

w=3.05 rad/s or 29.88rpm

Explanation:

k = coefficient of friction = 0.3900

R = radius of the cylinder = 2.7m

V = linear speed of rotation of the cylinder

w = angular speed = V/R or to rewrite V = w*R

N = normal force to cylinder

N= $=\frac{m(V)^{2}}{R}=m*(w)^2*R$

$Friction force\\Ff = k*N\\Ff= k*m*w^2*R$

$Gravitational force \\Fg = m*g$

These must be balanced (the net force on the people will be 0) so set them equal to each other.

$Fg = Ff$

$m*g = k*m*w^2*R$

$g=k*w^{2}*R$

$w^2 =\frac{g}{k*R}$

$w=\sqrt{\frac{g}{k*R}} \\w =\sqrt{\frac{9.8\frac{m}{s^{2}}}{0.3900*2.7m}}\\ w=\sqrt{9.306}=3.05 \frac{rad}{s}$

There are 2*pi radians in 1 revolution so:

$RPM=\frac{w}{2\pi }*60\\RPM=\frac{3.05\frac{rad}{s}}{2\pi}*60\\RPM= 0.498*60\\RPM=29.88$

So you need about 30 RPM to keep people from falling out the bottom

7 0

3 years ago

A vector must always have both size and ..

DochEvi [55]

Both magnitude and DIRECTION
For example,
• 12m East
• -2 miles
•9 meter north
• 8 miles up

5 0

2 years ago