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mamaluj [8]
3 years ago
12

The two parents shown in (Figure 1) exert upward forces

Physics
1 answer:
timama [110]3 years ago
5 0

Answer:

W = 83 [N] (weight)

Explanation:

We must clarify that the system is in static equilibrium, in such a way that we must perform a sum of forces on the Y-axis equal to zero, in order to determine the downward force corresponding to the weight of the child.

∑Fy = 0

F_{1}+F_{2}-m*g=0\\18+65=m*g\\m*g = 83 [N]

And the mass can be determined as follows:

m*g=83\\m = 83/9.81\\m=8.46[kg]

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A speeding motorist traveling 120 km/h passes a stationary police officer. The officer immediately begins pursuit at a constant
Lelu [443]
120 km/h = 33.33 m/s   10.8 km/h/s = 3 m/s/s     the motorist will cover a distance of 33.33 m x T(seconds)   the police officer will cover a distance of 1/2 (3) T^2   they will be at the same point when   33.33T = 1.5 T^2   33.33 = 1.5 T   3T = 66.66   T = 22.22 seconds   it will take the officer 22.22 seconds to catch the motorist.     the officer will be moving V=AT = 3m/s x 22.22 seconds = 66.66 m/s   almost 240 km/h (239.976 km/h)
8 0
4 years ago
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(a) How fast must a 3000-kg elephant move to have the same kinetic energy as a 65.0-kg sprinter running at 10.0 m/ s?
Aleksandr-060686 [28]

Answer:

1.4719 m per sec

Explanation:

Hello

Kinetic energy is the energy associated with the movement of objects. Although there are many forms of kinetic energy  

the formula to use is

E=\frac{mv^{2} }{2}

where m is the mass of the object and v the velocity

lets see the kinetic energy of the sprinter running

E=\frac{65 Kg*10(\frac{m}{s} ^)){2} }{2} \\\\E=\frac{65 *100 }{2} \\E=3250 Joules\\\\

Now, the elephant must have the same kinetic energy

E=\frac{m*v_{2} ^{2} }{2} \\\\E*2=m*v_{2} ^{2}\\ \frac{2E}{m} =v_{2} ^{2} \\\sqrt{\frac{2E}{m} } =v_{2}  \\\\\\v_{2} =\sqrt{\frac{2*3250}{3000} }\\ \\v_{2} =1.4719 \frac{m}{s} \\\\

it works only the positive root, so the elephant must to  walk  to 1.4719 m/s to have the same kinetic energy.

Have a great day

8 0
3 years ago
Manny walked a total of 3 miles. The reference point used to calculate the total distance that he walked was the same as the end
yawa3891 [41]

Correct question:

Manny walked a total of 3 miles. The reference point used to calculate the total distance that he walked was the same as the ending point. Which describes where Manny most likely walked?

a. from the bottom of a hill to the top

b. on a circular nature trail

c. on a sidewalk from his house to the mall

d. from the beginning of a straight track to the end

Answer:

b. on a circular nature trail

Explanation:

As it is mentioned that Manny that his reference point from where she started is same as ending point meaning that she moved in a circular path making point B correct answer.

7 0
3 years ago
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Sometimes a new experimental result may seem to go against a well-established scientific theory. Which statement BEST describes
MrMuchimi
The best thing to do in this case is to redo the experiment and re record the info, it has to be precise and accurate so you also have to check if your procedure is correct. If the results are both accurate and precise then you have to report your findings to the committee of that specific field. <span />
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3 years ago
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A circular wire loop of radius LaTeX: RR lies in the xy-plane with the z-axis running through its center. There is initially no
frosja888 [35]

Answer:

Explanation:

Given a circular loop of radius R

r = R

Note: the radius lies in the xy plane

Area is given as

A = πr² = πR²

At t = 0, no magnetic field B=0

The magnetic field is given as a function of time

B = C•exp(t) •i + D•t² •k

Where C and D are constant

We want to find the magnitude of EMF in the circular loop.

EMF is given as

ε = - N•dΦ/dt

Where,

N is number of turn and in this case we will assume N = 1.

Φ is magnetic flux and it is given as

Φ = BA

ε = - N•d(BA)/dt

Where A is a constant, then we have

ε = - N•A•dB/dt

B = C•exp(t) •i + D•t² •k

dB/dt = C•exp(t) •i + 2D•t •k

Then,

ε = - N•A•dB/dt

ε = - 1•πR²•(C•exp(t) •i + 2D•t •k)

ε = -πR²•(C•exp(t) •i + 2D•t •k)

So, let find the magnitude of EMF

Generally finding magnitude of two vectors R = a•i + b•j

Then, |R| = √a² + b²

So, applying this we have,

ε = πR² (√(C²•exp(2t) + 4D²t²))

From the given magnetic field, we are given that,

B = 0 at t = 0

B = C•exp(t) •i + D•t² •k

B = 0 = C•exp(0) •i + D•0² •k

0 = C

Then, C = 0.

So, substituting this into the EMF.

ε = πR² (√(0²•exp(2t) + 4D²t²))

ε = πR² (√4D²t²)

ε = πR² × 2Dt

ε = 2πDR²t

So, the EMF is also a function of time

ε = 2πDR²t

4 0
4 years ago
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