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motikmotik
3 years ago
10

7. A wrench falls from a helicopter that is rising steadily at +6.0 m/s.

Physics
1 answer:
Levart [38]3 years ago
7 0

A. The initial velocity of the wrench is + 6 m/s

B. The velocity of the wrench at two seconds is 13.6 m/s downward

C. The wrench is at 7.6 meters from the release point after 2 seconds

D. The wrench is 19.6 meters below the helicopter after 2 seconds

Explanation:

A wrench falls from a helicopter that is rising steadily at +6.0 m/s

1. The speed of the helicopter is 6 m/s

2. The initial speed of the wrench is the same speed of the helicopter

A.

The initial velocity of the wrench is  + 6 m/s

B.

The velocity after 2 seconds can be get by using the rule:

→ v = u + gt

where v is the final velocity u is the initial velocity, g is the

acceleration of gravity and t is the time

→ u = 6 m/s , t = 2 seconds , g = -9.8 m/s²

→ v = 6 - 9.8(2) = -13.6 m/s

Negative sign means the direction of velocity is downward

The velocity of the wrench at two seconds is 13.6 m/s downward

C.

We can find the distance that the wrench releases by using the rule:

→ s = ut + \frac{1}{2} g t²

where s is the distance

→ u = 6 m/s , g = -9.8 m/s² , t = 2

→ s = 6(2) - \frac{1}{2} (9.8)(2)²

→ s = 12 - 19.6 = -7.6 meters

We will ignore the negative sign because distance is a scalar quantity

The wrench is at 7.6 meters from the release point after 2 seconds

D.

The distance covered by the helicopter is the speed of the helicopter

Multiplied by the time

→ The speed of the helicopter is 6 m/s

→ The time is 2 seconds

→ The distance covered by the helicopter = 6 × 2 = 12 meters

→ The distance covered by the wrench is 7.6 meters

The total distance is the sum of the distance covered by the

helicopter and the distance covered by the wrench

→ The total distance = 7.6 + 12 = 19.6 meters

The wrench is 19.6 meters below the helicopter after 2 seconds

Learn more:

You can learn more about the free fall motion in brainly.com/question/5531630

#LearnwithBrainly

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a flag of mass 2.5 kg is supported by a single rope. A strong horizontal wind exerts a force of 12 N on the flag. Calculate the
tatuchka [14]
The free-body diagram of the forces acting on the flag is in the picture in attachment.

We have: the weight, downward, with magnitude
W=mg = (2.5 kg)(9.81 m/s^2)=24.5 N
the force of the wind F, acting horizontally, with intensity
F=12 N
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
T \cos \alpha -F=0
T \sin \alpha -W=
By dividing the second equation by the first one, we get
\tan \alpha =  \frac{W}{F}= \frac{24.5 N}{12 N}=2.04
From which we find
\alpha = 63.8 ^{\circ}
which is the angle of the rope with respect to the horizontal.

By replacing this value into the first equation, we can also find the tension of the rope:
T= \frac{F}{\cos \alpha}= \frac{12 N}{\cos 63.8^{\circ}}=27.2 N




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3 years ago
What is the weight of a 1-kilogram brick resting on a table?
MakcuM [25]

Answer:

The weight if the block is 10Newtons

Explanation:

The weight of any object is quantity of matter the object contains and it is always acting downwards on such body. This shows that the object is under the influence of gravity.

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W = mg

Give the mass of the brick to be 1kg

g is the acceleration due to gravity = 10m/s²

Weight of the object = 1 × 10

= 10kgm/s² or 10Newtons

5 0
2 years ago
Secondary evidence is the basis for drawing scientific conclusions is this true ir false
Leto [7]
The statement: "secondary evidence is the basis for drawing scientific conclusions" is definitely false. Secondary evidence is the body of information obtained to prove the existence of unknown or missing primary evidence. In drawing scientific conclusions, primary evidence is needed.
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2 years ago
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If the acceleration of a motorboat is 4.0 m/s2, and the motorboat starts from rest, what is its velocity after 6.0 s?
umka2103 [35]

Answer:

The velocity of the motorboat after 6s is 24 m/s.

Explanation:

Given;

acceleration of the motorboat, a = 4.0 m/s²

initial velocity of the motorboat, u = 0

time of motion of the motorboat = 6s

Apply the following kinematic equation to determine the velocity of the motorboat after 6 ;

v = u + at

v = 0 + (4 x 6)

v = 24 m/s

Therefore, the velocity of the motorboat after 6s is 24 m/s.

6 0
2 years ago
The flywheel of a steam engine runs with a constant angular velocity of 140 rev/min. When steam is shut off, the friction of the
ratelena [41]

Answer:

A) α = -1.228 rev/min²

B) 7980 revolutions

C) α_t = -8.57 x 10^(-4) m/s²

D) α = 21.5 m/s²

Explanation:

A) Using first equation of motion, we have;

ω = ω_o + αt

Where,

ω_o is initial angular velocity

α is angular acceleration

t is time the flywheel take to slow down to rest.

We are given, ω_o = 140 rev/min ; t = 1.9 hours = 1.9 x 60 seconds = 114 s ; ω = 0 rev/min

Thus,

0 = 140 + 114α

α = -140/114

α = -1.228 rev/min²

B) the number of revolutions would be given by the equation of motion;

S = (ω_o)t + (1/2)αt²

S = 140(114) - (1/2)(1.228)(114)²

S ≈ 7980 revolutions

C) we want to find tangential component of the velocity with r = 40cm = 0.4m

We will need to convert the angular acceleration to rad/s²

Thus,

α = -1.228 x (2π/60²) = - 0.0021433 rad/s²

Now, formula for tangential acceleration is;

α_t = α x r

α_t = - 0.0021433 x 0.4

α_t = -8.57 x 10^(-4) m/s²

D) we are told that the angular velocity is now 70 rev/min.

Let's convert it to rad/s;

ω = 70 x (2π/60) = 7.33 rad/s

So, radial angular acceleration is;

α_r = ω²r = 7.33² x 0.4

α_r = 21.49 m/s²

Thus, magnitude of total linear acceleration is;

α = √((α_t)² + (α_r)²)

α = √((-8.57 x 10^(-4))² + (21.49)²)

α = √461.82

α = 21.5 m/s²

4 0
2 years ago
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