I need help. It's 9th grade physical science

the coefficient of kinetic friction between a couch and the floor is 0.65. if the couch has a mass of 42 kg and you push it with

Anna [14]

Explanation:

The net force along the horizontal direction is

$\sum{F_x} = F_{applied} - f$

where f is the frictional force. We can find the frictional force by looking at the vertical forces acting on the couch:

$\sum{F_y} = N - mg = 0 \Rightarrow N = mg$

From the definition of frictional force,

$f = \mu{N} = \mu{mg} = (0.65)(42\:\text{kg})(9.8\:\text{m/s}^2)$

$\:\:\:\:= 267.5\:\text{N}$

Therefore, the net force on the couch is

$\sum{F_x} = 600\:\text{N} - 267.5\:\text{kN} = 332.5\:\text{N}$

4 0

1 year ago

Consider a large 1.54 V carbon-zinc dry cell used in a physics lab to supply 2.15 A to a circuit. The internal resistance of the

andrew-mc [135]

<h2>Answer:</h2>

1.77V

<h2>Explanation:</h2>

The electromotive force voltage (E) in a cell, is related to the lost voltage ( $V_{L}$ ) and the terminal voltage ( $V_{T}$ ) as follows;

E = $V_{T}$ - $V_{L}$

Where;

The lost voltage ( $V_{L}$ ) is the product of the internal resistance (r) of the cell and current (I) in the cell. i.e

$V_{L}$ = I x r

<em>Substitute </em> $V_{L}$ <em> = I x r into equation (i) as follows;</em>

E = $V_{T}$ - (I x r) ----------------------(ii)

<em>According to the question;</em>

E = 1.54V

I = 2.15A

r = 0.105Ω

<em>Substitute these values into equation(ii) as follows;</em>

1.54 = $V_{T}$ - (2.15 x 0.105)

1.54 = $V_{T}$ - (0.22575)

1.54 = $V_{T}$ - 0.22575

<em>Solve for </em> $V_{T}$ <em>;</em>

$V_{T}$ = 1.54 + 0.22575

$V_{T}$ = 1.77V

Therefore, the terminal voltage of the cell is 1.77V

8 0

2 years ago

An airplane has a takeoff speed of 62 m/s. assuming a constant acceleration of 1.7 m/s2, what is the minimum length of runway it

UkoKoshka [18]

Data:

u=0 m/s is the initial velocity of the plane

v=62 m/s is the final velocity of the plane (at which the plane takes off)

a=1.7 m/s^2 is the acceleration of the plane

To find the minimum distance S the plane needs to take off, we can use the following equation:

$2aS=v^2 -u^2$

Re-arranging it and substituting the numbers, we find

$S=\frac{v^2-u^2}{2a}=\frac{(62 m/s)^2-0}{2(1.7 m/s^2)}=1131 m$

3 0

2 years ago

A rocket accelerates by burning its onboard fuel, so its mass decreases with time. Suppose the initial mass of the rocket at lif

Serggg [28]

Answer:

Height of the rocket be one minute after liftoff is 40.1382 km.

Explanation:

$v(t)=-gt-v_e\times \ln \frac{m-rt}{m}$

v = velocity of rocket at time t

g = Acceleration due to gravity = $9.8 m/s^2$

$v_e$ = Constant velocity relative to the rocket = 2,900m/s.

m = Initial mass of the rocket at liftoff = 29000 kg

r = Rate at which fuel is consumed = 170 kg/s

Velocity of the rocket after 1 minute of the liftoff =v

t = 1 minute = 60 seconds'

Substituting all the given values in in the given equation:

$v(60)=-9.8 m/s^2\times 60 s-2,900m/s\times \ln (\frac{29,000 kg-170 kg/s\times 60 s}{2,9000 kg})$

$v(60) = 668.97 m/s$

Height of the rocket = h

$Velocity=\frac{Displacement}{time}$

$668.97 m/s=\frac{h}{60 s}$

$h=668.97 m/s\times 60 s=40,138.2 m = 40.1382 km$

Height of the rocket be one minute after liftoff is 40.1382 km.

4 0

2 years ago

A. What are the three longest wavelengths for standing waves on a 240-cm-long string that is fixed at both ends?

vovangra [49]

Answer:

a) the three longest wavelengths = 4.8m, 2.4m, 1.6m

b) what is the frequency of the third-longest wavelength = 75Hz

Explanation:

The steps and appropriate formula and substitution is as shown in the attached file.

5 0

2 years ago