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serious [3.7K]
3 years ago
6

A 10 kg box hangs from a rope. What is the tension in the rope (in Newtons) if the box is stationary

Physics
1 answer:
Archy [21]3 years ago
5 0

Answer:

T = 98 N

Explanation:

The gravity of the earth is known to be 9.8 m/s²

Data:

  • m = 10 kg
  • g = 9.8 m/s²
  • T = ?

Use formula:

  • \boxed{\bold{T=m*g}}

Replace and solve:

  • \boxed{\bold{T=10\ kg*9.8\frac{m}{s^{2}}}}
  • \boxed{\boxed{\bold{T=98\ N}}}

The tension in the rope is <u>98 Newtons.</u>

Greetings.

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6. An earthquake releases two types of traveling seismic waves, called transverse and longitudinal waves. The average speed of t
zubka84 [21]

Answer:

The distance away the center of the earthquake is 1083.24 km.

Explanation:

Given that,

Speed of transverse wave = 9.1\ km/s

Speed of longitudinal wave = 5.7 km/s

Time = 71 sec

We need to calculate the distance of transverse wave

Using formula of distance

d=v\times t

d=9.1\times t....(I)

The distance of longitudinal wave

d=5.7\times (t+71)....(II)

From the first equation

t=\dfrac{d}{9.1}

Put the value of t in equation (II)

d =5.7\times(\dfrac{d}{9.1}+71)

\dfrac{9.1d-5.7d}{9.1}=71\times5.7

d0.3736=404.7

d =1083.24\ km

Hence, The distance away the center of the earthquake is 1083.24 km.

6 0
3 years ago
Which technique is best for manual in-line stabilization of a person floating faceup on the surface?
zmey [24]

Answer:

vise grip

Explanation:

Manual in-line stabilization (MILS) of the cervical spine is a type of airway management when dealing with  patients in traumatic condition ..it is a means that is performed by grasping the mastoid process of the patient, so as to prevent the movement of the cervical column during intubation of the trachea

MLS provides a means of stability to the cervical column for a patient in trauma. During this technique, a patient is restricted from moving his or her cervical collar. The vise grip can be used for a patient with neck injury. The technique is used to roll a patient to face up to prevent further injuries.

7 0
3 years ago
Read 2 more answers
After a check up, a person now has a far point of 100 cm, but with good near point vision. He needs to wear a new pair of correc
lakkis [162]

Answer:

so his far point according to this pair of glass is 200 cm

Explanation:

power of old pair of corrective glasses is given as

P = -0.5 dioptre

now we have

f = \frac{1}{P}

f = -2 m

f = -200 cm

now we know that for normal vision the maximum distance of vision is for infinite distance

so by lens formula we have

\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}

\frac{1}{d_i} + 0 = \frac{1}{200}

d_i = 200 cm

so his far point according to this pair of glass is 200 cm

7 0
3 years ago
Which artist of the northern European Renaissance, shown in this self-portrait, was also a block printer and engraver?
kicyunya [14]
It's very hard to see the self-portrait, so I can't identify him.
6 0
3 years ago
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Can someone please help me with this physics question? I'm desperate!
Lelu [443]

Answer:

a) 2·√10 seconds

b) Linda should be approximately 30.6 meters

c) Jenny's speed at the 100-m mark is approximately 6.325 m/s

Explanation:

The speed with which Linda is running = 8.6 m/s

The point Jenny starts = The 80-m mark

The acceleration of Jenny = 1.0 m/s²

a) The time it takes Jenny to run from the 80-m mark to the 100-m mark, <em>t</em>, is given as follows

Δs = u·t + (1/2)·a·t²

Δs = Distance = 100-m - 80-m = 20-m

u = The initial velocity of Jenny = 0

a = Jenny's acceleration = 1.0 m/s²

∴ 20 = 0×t + (1/2) × 1 × t² = t²/2

20 = t²/2

t = √(20 × 2) = 2·√10

The time it takes Jenny to run from the 80-m mark to the 100-m mark = 2·√10 seconds

b) The distance Linda runs in t = 2·√10 seconds, d = v × t

Given that Linda's velocity, v = 8.6 m/s, we have;

d = 8.0 × 2·√10 = 16·√10

The distance Linda runs in t = 2·√10 seconds = 16·√10 meters ≈ 50.6 meters

Therefore, Linda should be approximately (50.6 - 20) meters = 30.6 meters behind Jenny when Jenny starts running

c) Jenny's speed at the 100 m mark is given as follows;

v = u + a·t

t = 2·√10 seconds, a = 1.0 m/s², u = 0

∴ v = 0×t + 1.0×2·√10 = 2·√10 ≈ 6.325

Jenny's speed at the 100-m mark ≈ 6.325 m/s

3 0
3 years ago
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