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Dmitry [639]
3 years ago
5

An automobile with 0.500 m radius tires travels 80,000 km before wearing them out. How many revolutions do the tires make, negle

cting any backing up and any change in radius due to wear
Physics
1 answer:
Zinaida [17]3 years ago
5 0

Answer:

n = 25,464,790.89 revolutions

The tires have made 25,464,790.89 revolutions

Explanation:

Given;

Radius of tires r = 0.5 m

Total distance travelled d = 80,000 km = 80,000,000 m

1 revolution = 2πr

Total distance d = number of revolutions n × 2πr

d = n×2πr

d = 2πnr

Making n the subject of formula;

n = d/2πr

Substituting the given values;

n = (80,000,000)/(2×π×0.5)

n = 25,464,790.89470 revolutions

n = 25,464,790.89 revolutions

The tires have made 25,464,790.89 revolutions

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A person can jump 1.5m on the earth. How high could the person jump on a planet having the twice the mass of the earth and twice
MrMuchimi
F=mg=Gm1m2/r^2
g=Gm2/r^2
g=2Gm2/(2r)^2=2Gm2/4r^2=Gm2/2r^2
So since there is half times the gravity on this unknown planet that has twice earth's mass and twice it's radius, then the person can jump twice as high. 1.5*2= 3m high

5 0
3 years ago
Read 2 more answers
In a compound microscope, the objective has a focal length of 1.0 cm, the eyepiece has a focal length of 2.0 cm, and the tube le
expeople1 [14]

Answer:

The  value  is   m \approx   310

Explanation:

From the question we are told that

     The  focal length of the objective is  f_o =  1.0 \ cm

    The  focal length of the eyepiece is  f_e  =  2.0 \  cm

    The  tube length is  L  =  25 \  cm

Generally the magnitude of the overall magnification is mathematically represented as

            m =  m_o  *  m_e

Where  m_o is the objective magnification which is mathematically represented as

        m_o  =  \frac{L}{f_o }

=>      m_o  =  \frac{25}{1 }

=>      m_o  =  25

m_e is the eyepiece magnification which is mathematically evaluated as

     m_e  =  \frac{L }{f_e }

     m_e  =  \frac{25 }{ 2}

      m_e  =  12.5 \  cm

So

    m =  25 * 12.5

     m \approx   310

6 0
3 years ago
in the primitive yo-yo apparatus (figure 1), you replace the solid cylinder with a hollow cylinder of mass m , outer radius r ,
kirza4 [7]

The magnitude of the downward acceleration of the hollow cylinder is 6m/s^2.

Z = I α

T.R =1/2 M ( R^{2} + (R/2)^{2} )α

T.R = 1/2M 5R^{2}/4 α

T = 5Ma/8

Mg - T = Ma

Mg -  5Ma/8 =  Ma

Mg= 5Ma/8 +  Ma = 13Ma / 8

acceleration = 8g/13 = 6 m/s^2

The rate at which an object's velocity with respect to time changes is called its acceleration. The direction of the net force imposed on an item determines its acceleration in relation to that force. According to Newton's Second Law, the magnitude of an object's acceleration is the result of two factors working together

The size of the net balance of all external forces acting on that item is directly proportional to the magnitude of this net resultant force; the magnitude of that object's mass, depending on the materials from which it is built, is inversely related to its mass.

Learn more about acceleration here:

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4 0
1 year ago
A cart loaded with bricks has a total mass of 9.13 kg and is pulled at constant speed by a rope. The rope is inclined at 24.7 ◦
blagie [28]

Answer:

W = 0.63 KJ

Explanation:

Work (W) is defined as the point product of force (F) by the distance (d)the body travels due to this force.  

W= F*d *cosα Formula (1)  

F : force (N)

d : displacement (m)

α : angle between force and displacement

Newton's second law:

∑F = m*a Formula (2)  

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the cart on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the cart

W: Weight of the cart : In vertical direction

FN : Normal force : perpendicular to the floor

f : Friction force: parallel to the floor

T : tension Force,  inclined at  θ=24.7° above the horizontal

Calculated of the W

W= m*g

W= 9.13 kg* 9.8 m/s² = 89.47 N

x-y components o the  tension force (T)

Tx = Tcosθ = T*cos 24.7° (N)

Ty = Tsin θ = T*sin 24.7°  (N)

Calculated of the FN  

We apply the formula (2)  

∑Fy = m*ay ay = 0  

FN +Ty- W = 0  

FN = W-Ty  

FN =  89.47-T*sin 24.7°

Calculated of the friction force (f)

f = μk*FN

f =(0.597)*(  89.47-T*sin 24.7° )

f= 53.41-0.249T

Calculated of the tension force of the rope (f)

We apply the formula (2) :

∑Fx = m*ax  ,  ax= 0 ,because the speed of the cart  is constant

Tx - f = 0

T*cos 24.7°-( 53.41 - 0.249T )= 0

T*cos 24.7° + 0.249T = 53.41

(1.1575)T = 53.41

T= (53.41) / (1.1575)

T= 46.14 N

Work done on the cart by the rope

We apply the formula (1)

W=T*d *cosα

W= (46.14 N)*(15.1 m) *(cos24.7)

W = 632.97 (N*m) = 632.97 (J)

W = 0.63 KJ

6 0
3 years ago
What’s the formula for work
padilas [110]

Answer:

Fd

Explanation:

Work is force times distance. If you push on an object really hard but it does not budge, you have still performed no work on it, because anything times zero is still zero.

6 0
2 years ago
Read 2 more answers
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