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Oksanka [162]
3 years ago
11

A blue line with 5 orange tick marks then one red tick mark then 4 orange tick marks. The number zero is above the red tick mark

.Assume each tick mark represents 1 cm.Calculate the total displacement from 0 if an object moves 3 cm to the left, then 7 cm to the right, and then 6 cm to the left.The object moves cm to the left.What is the total distance the object travels? cm
Physics
1 answer:
Digiron [165]3 years ago
3 0

Answer:

16 cm

Explanation:

Given that,

The object begins from 0 and moves 3cm towards left side followed by 7 cm towards the right and then, 6 cm towards the left side.

Let the x-axis to be the +ve and on the right side and -ve on the left

Thus, displacement would be:

= 0 -3 + 7 -6

= -2 cm

This implies that the object displaces 2cm towards the left.

While the total distance covered by the object equal to,

= 0cm + 3cm + 7cm + 6cm

= 16 cm

Thus, <u>16 cm</u> is the total distance.

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Calculate the mass of an object with a density of 102.5 g/mL and volume of 375 mL.
AveGali [126]

Answer:

38,437.5

Explanation:

Density(d)= 102.5g/ml

Volume (v)=375ml

Mass(m) = ?

D =m/v

102.5= m/375

102.5*375=m

38,437.5=m

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6 0
4 years ago
A 0.50-kg mass attached to the end of a string swings in a vertical circle (radius 2.0 m). When the mass is at the highest point
il63 [147K]

Answer:

31.1 N

Explanation:

m = mass attached to string = 0.50 kg

r = radius of the vertical circle = 2.0 m

v = speed of the mass at the highest point = 12 m/s

T = force of the string on the mass attached.

At the highest point, force equation is given as

T + mg =\frac{mv^{2}}{r}

Inserting the values

T + (0.50)(9.8) =\frac{(0.50)(12)^{2}}{2}

T = 31.1 N

7 0
3 years ago
Read 2 more answers
In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400
liq [111]

Complete Question:

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Answer:

a) Speed of car A at the start of sliding = 4.23 m/s

b) speed of car B at the start of sliding = 3.957 m/s

c) Speed of car B before the collision = 7.28 m/s

Explanation:

NB: The figure is not provided but all the parameters needed to solve the question have been given.

Let the frictional force acting on car A, f_{ra} = \mu mg\\............(1)

Since frictional force is a type of force, we are safe to say f_{ra} = ma.......(2)

Equating (1) and (2)

ma = \mu mg\\a = \mu g\\\mu = 0.15\\a = 0.15 * 9.8 = 1.47 m/s^{2}

a) Speed of A at the start of the sliding

d_{A} = 6.1 m\\Speed of A at the start of sliding, v_{A} = \sqrt{2ad_{A} }\\ v_{A} = \sqrt{2*1.47*6.1 } \\v_{A} = \sqrt{17.934 } \\v_{A} = 4.23 m/s

b) Speed of B at the start of the sliding

d_{A} = 4.4 m\\Speed of A at the start of sliding, v_{B} = \sqrt{2ad_{B} }\\ v_{B} = \sqrt{2*1.47*4.4 } \\v_{B} = \sqrt{12.936 } \\v_{B} = 3.957 m/s

Let the speed of car B before collision = v_{B1}

Momentum of car B before collision = m_{B} v_{B1}

Momentum after collision = m_{A} v_{A} + m_{B} v_{B2}

Applying the law of conservation of momentum:

m_{B} v_{B1}  = m_{A} v_{A} +m_{B} v_{B2}

m_{A} = 1100 kg\\m_{B} = 1400 kg

(1400*v_{B1} ) = (1100 * 4.23) + ( 1400 * 3.957)\\(1400*v_{B1} ) = 10192.8\\v_{B1} = 10192.8/1400\\v_{B1 = 7.28 m/s

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4 years ago
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The answer is B


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