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Tpy6a [65]
2 years ago
7

A fast Humvee drove from desert A to desert B. For the first 12

Physics
1 answer:
Zina [86]2 years ago
4 0

Answer:

v = 172 km/h

Explanation:

For the first 12  hours, it traveled at an average speed of 185 km/h. Let d₁ is distance. So,

d_1=v_1\times t_1\\\\d_1=185\ km/h\times 12\ h\\\\d_1=2220\ km

For the  next 13 hours, it traveled at an average speed of 160 km/h. Let d₂ is the distance. So,

d_2=v_2\times t_2\\\\d_2=160\ km/h\times 13\ h\\\\d_2=2080\ km

Average speed = total distance/time taken

So,

v=\dfrac{d_1+d_2}{t_1+t_2}\\\\v=\dfrac{2220+2080}{12+13}\\\\v=172\ km/h

So, the average speed of the whole journey is 172 km/h.

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7. A mother pushes her 9.5 kg baby in her 5kg baby carriage over the grass with a force of 110N @ an angle
jasenka [17]

Weight of the carriage =(m+M)g =142.1\ N

Normal force =Fsin(\theta) + W = 197.1\ N

Frictional force =\mu N=27.59\ N

Acceleration =4.66\ m\ s^{-2}

Explanation:

We have to look into the FBD of the carriage.

Horizontal forces and Vertical forces separately.

To calculate Weight we know that both the mass of the baby and the carriage will be added.

  • So Weight(W) =(m+M)\times g =(9.5+5)\ kg \times 9.8 =142.1\ Newton\ (N)

To calculate normal force we have to look upon the vertical component of forces, as Normal force is acting vertically.We have weight which is a downward force along with F_x, force of 110\ N acting vertically downward.Both are downward and Normal is upward so Normal force =Summation\ of\ both\ forces

  • Normal force (N) = Fsin(\theta)+W=110sin(30) + 142.1 =197.1\ N
  • Frictional force (f) =\mu N=0.14\times 197.1 =27.59\ N

To calculate acceleration we will use Newtons second law.

That is Force is product of mass and acceleration.

We can see in the diagram that F_y=Horizontal and F_x=Vertical component of forces.

So Fnet = Fy(Horizontal) - f(friction) = m\times a

  • Acceleration (a) =\frac{Fcos(\theta)-\mu N}{mass(m)} =\frac{(95.26-27.59)}{14.5}= 4.66\ m\ s{^2 }

So we have the weight of the carriage, normal force,frictional force and acceleration.

3 0
1 year ago
Power is __________________. Power is __________________. the work done by a system the force required to push something the spe
DaniilM [7]

Answer:

The rate at which the energy of a system is transformed

Explanation:

Power is the rate at which energy of a system is transformed or the rate at which work is done. It is defined by Power = Workdone/time taken

Its unit is the Watt denoted by the letter W.

For example, assuming a work of 200 J is done in 10 s, then Power, P equals

P = workdone/time taken = 200 J/10 s = 20 J/s = 20 W

6 0
1 year ago
Samantha wants to study circus performance when she gets to college. She has mastered many physical skills already, but she keep
Mekhanik [1.2K]
Power is what she should work on. 
8 0
2 years ago
Read 2 more answers
A box weighing 46 newtons rests on an incline that makes an angle of 25° with the horizontal. What is the magnitude of the compo
saul85 [17]

The magnitude of the component of the box’s weight perpendicular to the incline can be olve using the formula:

F = wcos(a)

Where F is the box’s weight perpendicular to the incline

W is the weight of the box

A is the angle of the incline

F = (46)cos(25)

F = 42 N

8 0
2 years ago
Two 7.0cm×7.0cm metal electrodes are spaced 1.0 mm apart and connected by wires to the terminals of a 9.0 v battery.
AleksandrR [38]

Answer:

q=390.285x10^{-12}C

Explanation:

Those kind of problems of electric physics is about capacitors, so the normal questions are:

What are the charge on each electrode?

Solve this you can get other information required in the problem or write down the other questions you need

7.0cm*\frac{1m}{100cm}=70x10^{-3}m

A=70x10^{-3}m*70x10^{-3}m=4.9x10^{-3}m62

Capacitance

C=E_o*\frac{A}{d}

E_o=8.85x10^{-12}\frac{C^2}{N*m^2}

C=8.85x10^{-12}\frac{C^2}{N*m^2}*\frac{4.9x10^{-3}m^2}{1x10^{-3}m}

C=43.365x10^{-12}F

C=43.365pF

The charge is find by the equation

q=C*V

q=43.365pF*9V

q=390.285x10^{-12}C

5 0
2 years ago
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