What is the speed of a person that walk 400 meters in 1900 seconds

A particle of mass 73 g and charge 67 µC is released from rest when it is 47 cm from a second particle of charge −25 µC. Determi

stich3 [128]

Answer:

933.804423995 m/s²

Explanation:

$q_1$ = Charge on particle 1 = 67 µC

$q_2$ = Charge on particle 2 = -25 µC

r = Distance between the particles = 47 cm

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

m = Mass of particle = 73 g

Electric force is given by

$F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^{9}\times -25\times 10^{-6}\times 67\times 10^{-6}}{0.47^2}\\\Rightarrow F=-68.1677229516\ N$

The magnitude of force is 68.1677229516 N

Acceleration is given by

$a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{68.1677229516}{0.073}\\\Rightarrow a=933.804423995\ m/s^2$

The acceleration is 933.804423995 m/s²

8 0

3 years ago

A system of pulleys is called an ?

lesantik [10]

A block and a tackle

3 0

3 years ago

A cat is sleeping on the floor in the middle of a 2.8-m-wide room when a barking dog enters with a speed of 1.40 m/s. As the dog

marysya [2.9K]

Answer:

the cat is 0.4238 m in front of the dog as it leaps through the window

Explanation:

Given that;

acceleration a = 0.85 m/s²

speed v = 1.40 m/s

the cat is at rest, so initial velocity u = 0

we know that, since the cat is sleeping on the floor in the middle of a 2.8-m-wide room, it needs to cover (2.8 m / 2 ) distance to get to the window;

using the second equation equation of motion;

s = ut + 1/2 at²

we substitute

2.8/2 = 0×t + 1/2 × 0.85 × t²

1.4 = 0.425t²

t = √( 1.4 / 0.425 )

t = 1.81497 sec

now, at acceleration 0.10 m/s²

the dog has to cover the distance;

s = ut + 1/2 at²

s = ( 1.4 × 1.81497) - 1/2 × 0.10 × 1.81497²

s = 2.540958 - 0.1647

s = 2.3762 m

The cant in front of the dog as it leaps through the window;

distance = 2.8 m - 2.3762 m

distance = 0.4238 m

Therefore, the cat is 0.4238 m in front of the dog as it leaps through the window

7 0

3 years ago

With that push, they have accelerated their masses, which means they have _____ of their bodies.

bixtya [17]

Answer: changed the velocity

Hope this helps lol :)

8 0

3 years ago

A ball thrown straight up into the air is found to be moving at 6.79 m/s after falling 1.87 m below its release point. Find the

MaRussiya [10]

Answer:

3.07 m/s

Explanation:

We know that from kinematics equation

$v^{2}=u^{2}+2as$ and here, a=g where v is the final velocity, u is the initial velocity, a is acceleration, s is the distance moved, g is acceleration due to gravity

Making u the subject then

$u=\sqrt {v^{2}-2gs}$

Substituting v for 6.79 m/s, s for 1.87 m and g as 9.81 m/s2 then

$u=\sqrt {6.79^{2}-(2\times 9.81\times 1.87)}=3.068338313 m/s\approx 3.07 m/s$

5 0

3 years ago