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Cloud [144]
3 years ago
13

Joe rides his bicycle in a straight line for 2 hour with an average velocity of 13 km/h south, how far has he ridden?

Physics
1 answer:
dmitriy555 [2]3 years ago
8 0

Answer:

ans is 3.125 km

i hope this is correct

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1. An object in free fall will have an initial velocity equal to zero when: a. It is thrown vertically down
Ne4ueva [31]

Answer:

b. It is dropped

Explanation:

If the initial velocity is zero, the object move from rest. That happens if the object is dropped

6 0
3 years ago
What is the difference between a 3 v and 12 v battery
Lorico [155]
V=IR
The more V, the more I
6 0
3 years ago
A 2,000 kg rocket is launched 12 km straight up at a constant acceleration into the sky at which point the rocket is travelling
hodyreva [135]

Answer:

797700000 J

Explanation:

From the question,

The work done by the rocket, is given as,

W = Ek+Ep............. Equation 1

Where Ek and Ep are the potential and the kinetic energy of the rocket respectively.

Ep = mgh............ Equation 2

Ek = 1/2mv²............. equation 3

Substitute equation 2 and equation 3 into equation 1

W = mgh+1/2mv².............. Equation 4

Where m = mass of the rocket, h = height, v = velocity of the rocket, g = acceleration due to gravity.

Given: m = 2000 kg, h = 12 km = 12000 m, v = 750 m/s, g = 9.8 m/s²

Substitute into equation 4

W = 2000(12000)(9.8)+1/2(2000)(750²)

W = 235200000+562500000

W = 797700000 J

4 0
3 years ago
A truck using a rope to tow a 2230-kg car accelerates from rest to 13.0 m/s in a time of 15.0s. How strong must the rope be? μk
Leokris [45]

Answer:

The rope must have a force of 10084,21 N

Explanation

Acceleration calculation

The car acceleration is equal to the acceleration of the truck

ac: car acceleration\frac{m}{s^{2} }

at: truck acceleration\frac{m}{s^{2} })

ac = at= \frac{vf-vi}{t-ti}  equation(1)

Known information:

vi = Initial speed = 0, ti = initial time = 0

vf = Final speed = 13 \frac{m}{s}, t = final time =5 s

We replaced the known information in the equation(1):

ac = at = \frac{13-0}{15-0}

ac=ac=\frac{13}{15}  \frac{m}{s}

Dynamic analysis

The forces acting on the car are the following:

Wc: Car weight

N: normal force, road force on the car

Ff: Friction force

T: Force of tension

Car weight calculation:

Wc=mc*g

mc = Car mass = 2230kg

g = Gravity acceleration=9.8 \frac{m}{s^{2} }

Wc= 2230*9.8

Wc=21854 N

Normal force calculation:

Newton's first law

sum Fy= 0

N-W=0

N=W

N=21854 N

Friction force calculation (Ff):

We have the formula to calculate the friction force:

Ff = μk * N  Equation (3)

μk kinetic coefficient of friction

We know that μk = 0.373and N= 21854N ,then:

Ff=0.373*21854

Ff=8151.54 N

Calculation of the tension force in the rope (T):

Newton's Second law

sum Fx= mc*ac

T-Ff=mc*ac

T=2230(\frac{13}{15}) + 8151.54

T=10084,21 N

Answer: The rope must have a force of 10084,21 N

8 0
3 years ago
Refrigerant-134a enters an adiabatic compressor as saturated vapor at -24C and leaves at 0.8 MPa and 60C. When the mass flow r
sergey [27]

Answer:

= 287kW

Explanation:

Knowing the enthalpy data, we have to

h1=239.16kJ/kg\\h2=296.81kJ/kg

So,

E_i_n_p_u_t=mh_1

Here,

m=mass flow rate

h= Enthalpy of refrigerant at the compressor

Replacing

= 1.2 × 239.16

= 287kW

4 0
3 years ago
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