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brilliants [131]
3 years ago
11

A spelunker is surveying a cave. She follows a passage 140 m straight west, then 290 m in a direction 45? east of south, and the

n 280 m at 30? east of north. After a fourth unmeasured displacement, she finds herself back where she started. Use a scale drawing to determine the magnitude of the fourth displacement. Determine the direction of the fourth displacement.
Physics
1 answer:
zimovet [89]3 years ago
3 0

Answer:

Magnitude of fourth displacement is approximately 95 metres,

Direction of fourth displacement is straight west.

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A square loop of wire lies in the x, y plane with opposite corners at (0, 0) and (a, a). The loop carries a current I directed t
Ket [755]

Answer:

Please refer to the figure.

Explanation:

The magnitude of the magnetic field can be found by Biot-Savart Law. We should divide the loop into four components. Each component has a similar solution but their directions are quite different.

The directions can be found by right-hand rule. Point your index finger into the direction of current, point your middle finger towards the target point (0,0,a). Your thumb will show you the direction of magnetic field.

5 0
3 years ago
Two forces of 50 N and 30N respectively, are acting on an object. Find the net force (in N) on the object if
xxTIMURxx [149]

A) the forces are acting in the same direction.. B) Together, forces are acting in opposite directions

Answer:

A) 80 N

B) 20 N

Explanation:

A) If the forces acting are in the same direction, then the net force will be a sum of both so many faces..

Thus;

ΣF = 50 + 30

ΣF = 80 N

B) If the forces are acting in the in opposite directions with the larger force pointing in the positive y-axis then, the net force is;

ΣF = 50 - 30

ΣF = 20 N

8 0
3 years ago
How to do this, i'm completely lost
vaieri [72.5K]
There are two torques t1 and t2 on the beam due to the weights, one torque t3 due to the weight of the beam, and one torque t4 due to the string.

You need to figure out t4 to know the tension in the string.

Since the whole thing is not moving t1 + t2 + t3 = t4.

torque t = r * F * sinФ = distance from axis of rotation * force * sin (∡ between r and F)

t1 =3.2 * 44g 
t2 = 7 * 49g 
t3 = 3.5 * 24g 

t4 = t1 + t2 + t3 = 5570,118

The t4 also is given by:

t4 = r * T * sin Ф

r = 7
Ф = 32°
T: tension in the string

T = t4 / (r * sinФ)

T = t4 / (7 * sin(32°)) 

T = 1501,6 N

8 0
3 years ago
which of the folloing statements about ionization energy is true? A elements toward the bottom of a group periodic table general
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<span>
accept the flow of electrons.resist the flow of electrons.accept the flow of protons.resist the flow of protons.It is one of these </span>
6 0
3 years ago
An astronaut finds herself in a predicament in which she has become untethered from her shuttle. She figures that she could get
Blizzard [7]

In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.

The equation that defines the linear moment is given by

mV_i = (m-m_O)V_f - m_OV_O

where,

m=Total mass

m_O = Mass of Object

V_i = Velocity before throwing

V_f = Final Velocity

V_O = Velocity of Object

Our values are:

m_1=5.3kgm_2=7.9kg\\m_3=10.5kg\\m_A=75kg\\m_{Total}=m=98.7Kg

Solving to find the final speed, after throwing the object we have

V_f=\frac{mV_0+m_TV_O}{m-m_O}

We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.

That way during each section the equations should be modified depending on the previous one, let's start:

A) 5.3Kg\rightarrow 15m/s

V_{f1}=\frac{mV_0+m_TV_O}{m-m_O}

V_{f1}=\frac{(98.7)*0+5.3*15}{98.7-5.3}

V_{f1}=0.8511m/s

B) 7.9Kg\rightarrow 11.2m/s

V_{f2}=\frac{mV_{f1}+m_TV_O}{m-m_O}

V_{f2}=\frac{(98.7)(0.8511)+(7.9)(11.2)}{98.7-5.3-7.9}

V_{f2} = 2.0173m/s

C) 10.5Kg\rightarrow 7m/s

V_{f3}=\frac{mV_{f2}+m_TV_O}{m-m_O}

V_{f3}=\frac{(98.7)(2.0173)+(10.5)(7)}{98.7-5.3-7.9-10.5}

V_{f3} = 3.63478m/s

Therefore the final velocity of astronaut is 3.63m/s

7 0
3 years ago
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