Ask question

Ask question

All categories

2 years ago

11

A 5 kg block accelerates across a horizontal surface at 7 m/s^2 when you exert a 60 N force on it. What is the force of friction

on the block (Ffs/b)?
If you don’t mind finding the kinetic friction coefficient too?

1 answer:

alina1380 [7]2 years ago

3 0

5 kg block and 7 m/s^2 = 14

You might be interested in

Needing help with this please

Gemiola [76]

Answer:

So A we cant sadly do because we cant draw. B is going to be kinetic. Thats because static friction means it stays in one place, for kinetic it means moving. So it will be 0.05 as the coefficient of the friction. Sadly, I cannot calculate C. You will have to use trigonemetry but I cannot fit that big an explanation.

Answer to A: the free body diagram would be the ski things inclined with gravity, friction, and air resistance. I except you know which directions

Answer to B: Kinetic friction is the answer.

Answer to C: Find on own, I cannot write super big explanations - use trigonometry.

6 0

2 years ago

When you go "down" the food chain by continuing to ask "what does it eat?" at what category of living things do you always end u

Klio2033 [76]

Answer:

Autotrophs

Explanation:

When you go down a food chain continuing to ask "what does it eat?" the last living thing that you will land upon is an autotroph.

Autotrophs are the primary producers as they (photoautotrophs) use the energy either from the sun to prepare there food by the process of photosynthesis or, more rarely, obtain chemical energy through oxidation (chemoautotrophs) to make organic substances from inorganic ones.

Autotrophs get consumed by the primary consumers in the food chain.

7 0

2 years ago

umka2103 [35]

Answer:

$x=2.4365\ m$

and

$x=-1.4365\ m$

Explanation:

Given:

first charge, $q_1=5\times 10^{-3}\ C$

second charge, $q_2=3\times 10^{-3}\ C$

position of first charge, $x_1=-2\ m$

position of second charge, $x_2=-1\ m$

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

$E_1=E_2$

since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

$\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}$

$\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}$

$3(r^2+1+2r)=5r^2$

$2r^2-6r-3=0$

$r=3.4365 \&\ r=-0.4365$

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

$x=-1+3.4365=2.4365\ m$

and

$x=-1-0.4365=-1.4365\ m$

6 0

2 years ago

A college student holds a pail full of water by the handle and whirls it around in a vertical circle at a constant speed. The ra

Pavlova-9 [17]

The minimum speed of the water must be 3.4 m/s

Explanation:

There are two forces acting on the water in the pail when it is at the top of its circular motion:

The force of gravity, mg, acting downward (where m is the mass of the water and g the acceleration of gravity)
The normal reaction, N also acting downward

Since the water is in circular motion, the net force must be equal to the centripetal force, so:

$N+mg=m\frac{v^2}{r}$

Where:

$g=9.8 m/s^2$

v is the speed of the pail

r = 1.2 m is the radius of the circle

The water starts to spill out when the normal reaction of the pail becomes zero:

N = 0

When this occurs, the equation becomes:

$mg=m\frac{v^2}{r}\\v=\sqrt{gr}$

And substitutin the values of g and r, we find the minimum speed that the water must have in order not to spill out:

$v=\sqrt{(9.8)(1.2)}=3.4 m/s$

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

6 0

2 years ago

A celebrated Mark Twain story has motivated contestants in the Calaveras County Jumping Frog Jubilee, where frog jumps as long a

IrinaK [193]

The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

To find the answer, we need to know about the time of flight and range of projectile motion.

<h3>What's the expression of range of a projectile motion?</h3>

Range = U²× sin(2θ)/g
U= initial velocity, θ= angle of projectile and g= acceleration due to gravity
U=√{Range×g/sin(2θ)}
Here, range= 2.20m, = 36.5°
U= √{2.20×9.8/sin(73)}

U= √{2.20×9.8/sin(73)} = 22.5m/s

<h3>What's the expression of time of flight in projectile motion?</h3>

Time of flight= (2×U×sinθ)/g
So, T= (2×22.5×sin36.5°)/9.8

= 2.73 s

Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

Learn more about the range and time period of projectile motion here:

brainly.com/question/24136952

#SPJ1

4 0

1 year ago