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luda_lava [24]
3 years ago
12

Mary starts from her house, walks 80 meters south, and stops to chat with her aunt on the sidewalk. After chatting for a few min

utes, she walks 125 meters to the west to meet a friend and then walks 45 meters north with her friend to a cafe for lunch. Overall, she takes 10 minutes to do all this. What is Mary’s average speed?Select one of the options below as your answer: A. 8 meters/second B. 1.25 meters/second C. 25 meters/second D. 4.5 meters/second E. 0.42 meters/second
Physics
1 answer:
marin [14]3 years ago
5 0
Mary walks:
d 1 = 80 m,  d 2 = 125 m,  d 3 = 45 m
t = 10 minutes = 600 seconds;
Average speed:
v = ( d 1 + d 2 + d 3 ) / t 
v = ( 80 m + 125 m + 45 m ) / 600 s
v = 250 m / 600 s
v = 0.4167 m/s ≈ 0.42 m/s
Answer:
E ) 0.42 meters/second
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Explanation:

There are many ways to achieve a healthier lifestyle and increase fitness. Healthy eating and constant exercise are essential for maintaining good physical and mental health. Through effective nutrition the human body is able to function better, there is an increase in disposition, improvement in immunity, improvement in blood rates, etc. Physical exercise, on the other hand, assists in strengthening muscles, controlling blood pressure, preventing cardiovascular disease, greater satisfaction, less stress, etc.

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6 0
3 years ago
An engine is used to lift a beam weighing 9,800 N up to 145 m. How much work must the engine do to lift this beam? How much work
Arturiano [62]

Explanation:

Given that,

Weight of the engine used to lift a beam, W = 9800 N

Distance, d = 145 m

Work done by the engine to lift the beam is given by :

W = F d

W=9800\ N\times 145\ m\\\\W=1421000\ J\\\\W=1421\ kJ

Let W' is the work must be done to lift it 290 m. It is given by :

W'=9800\ N\times 290\ m\\\\W'=2842000\ J\\\\W'=2842\ kJ

Hence, this is the required solution.

5 0
3 years ago
A 25.0-g sample of copper at 363 K is placed in I 00.0 g of water at 293 K. The copper and water quickly come to the sa me tempe
Simora [160]

Answer:

Final temperature is 295K

Explanation:

Where the sample of copper is placed in the water, the heat transferred from the copper is equal that the heat absorbed by the water.

The heat transferred from the copper is:

C×\frac{1mol}{63,546g}×mass×ΔT

Where C is molar heat capacity of copper (24,5J/molK)

Mass is 25,0g

And ΔT is final temperature - initial temperature (X-363K)

Also, the heat absorbed by the water is:

-C×\frac{1mol}{18,02g}×mass×ΔT

Where C is molar heat capacity of water (75,2J/molK)

Mass is 100,0g

And ΔT is final temperature - initial temperature (X-293K)

As heat transferred is equal to heat absorbed:

24,5J/molK×\frac{1mol}{63,546g}×25,0g×(X-363K) = -75,2J/molK×\frac{1mol}{18,02g}×100,0g× (X-293K)

9,64X J/K - 3499J = - 417X J/K + 122273J

426,64X J/K = 125772 J

<em>X = 295K</em>

<em></em>

Final temperature is 295K

I hope it helps!

6 0
3 years ago
Four equal masses m are so small they can be treated as points, and they are equallyspaced along a long, stiff mass less wire. T
gavmur [86]

The moment of inertia of a point mass about an arbitrary point is given by:

I = mr²

I is the moment of inertia

m is the mass

r is the distance between the arbitrary point and the point mass

The center of mass of the system is located halfway between the 2 inner masses, therefore two masses lie ℓ/2 away from the center and the outer two masses lie 3ℓ/2 away from the center.

The total moment of inertia of the system is the sum of the moments of each mass, i.e.

I = ∑mr²

The moment of inertia of each of the two inner masses is

I = m(ℓ/2)² = mℓ²/4

The moment of inertia of each of the two outer masses is

I = m(3ℓ/2)² = 9mℓ²/4

The total moment of inertia of the system is

I = 2[mℓ²/4]+2[9mℓ²/4]

I = mℓ²/2+9mℓ²/2

I = 10mℓ²/2

I = 5mℓ²

4 0
3 years ago
A 53.0-kg skater is traveling due east at a speed of 2.90 m/s. A 72.0-kg skater is moving due south at a speed of 6.20 m/s. They
soldier1979 [14.2K]

Answer:

a.\thta=71^{\circ}

b.v_f=3.78 m/s

Explanation:

We are given that

m_1=53 kg

v_1=2.9 m/s

m_2= 72 kg

v_2=6.2 m/s

a.We have to find the angle

\theta=tan^{-1}{\frac{m_2v_2}{m_1v_1}=\frac{72(6.2)}{53(2.9)}=71^{\circ}

\theta=71^{\circ}

b. We have to find the speed v_f

According to law of conservation of momentum

m_1v_1=(m_1+m_2)v_fcos\theta

53(2.9)=(53+72)v_fcos 71^{\circ}=40.7 v_f

v_f=\frac{153.7}{40.7}=3.78 m/s

3 0
3 years ago
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