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zimovet [89]
3 years ago
14

Solve the inequality 2(n+3) – 4<6. Then graph the solution.

Physics
1 answer:
Aloiza [94]3 years ago
6 0
The solution is 22 2(n+3)-4&6
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Explain why, in terms of forces, there is a risk of head injury when diving from this height. Suggest why the high divers would
podryga [215]

Answer:

Due to lower risk of injury or damage.

Explanation:

The high divers would choose to enter the water from the feet first because there is low risk of injury. The brain is the most important part of the body which very sensitive to any small injury. Small injury to brain leads to big problems in life. High divers can reach speeds of nearly 60 mph and enters about 28m into the water in about three seconds which can damage the head region if comes in contact with the ground so this is the reason the high divers avoid of entering in the water through their heads and choose entering through their feet.

4 0
3 years ago
A turtle ambles leisurely, as turtles tend to do, when it moves from a location with position vector 1,=1.91 m and 1,=−2.73 m in
Elena-2011 [213]

Answer:

Components: 0.0057, -0.0068. Magnitude: 0.0089 m/s

Explanation:

The displacement in the x-direction is:

d_x = 3.65-1.91=1.74 m

While the displacement in the y-direction is:

d_y = -4.79 -(-2.73)=-2.06 m

The time taken is t = 304 s.

So the components of the average velocity are:

v_x = \frac{d_x}{t}=\frac{1.74}{304}=0.0057 m/s

v_y = \frac{d_y}{t}=\frac{-2.06}{304}=-0.0068 m/s

And the magnitude of the average velocity is

v=\sqrt{v_x^2+v_y^2}=\sqrt{(0.0057)^2+(-0.0068)^2}=0.0089 m/s

8 0
3 years ago
What<br> must always be<br> included on the<br> graph
Nookie1986 [14]
Clearly visible data points and appropriate labels on each access that include units
4 0
3 years ago
An early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary
brilliants [131]

Answer:

Explanation:

When the craft was stationary , weight will be balanced by tension

T = mg

T = 7000 N

A)

when the craft was being lowered to the seafloor

drag force will act in upper direction , so

T₁ + 1800 = mg

T₁  = mg - 1800

= 7000 - 1800

= 5200 N

52 X 10² N

B)

when the craft was being raised from the seafloor , Tension will act in downward direction

T₂ = mg+ 1800

T₂  = 7000 - 1800

= 8800N

3 0
3 years ago
How long will be required for an object to go from a speed of 22m/s to a speed of 27m/s if the acceleration is 5.93m/s^2 ?
mario62 [17]

Answer:

Required time, t = 0.84 seconds

Explanation:

It is given that,

Initial speed of an object, u = 22 m/s

Final velocity of an object, v = 27 m/s

Acceleration, a = 5.93 m/s²

We have to find the time required for an object to go a speed of 22 m/s to a speed of 27 m/s. It can be solved by using first equation of motion as:

v=u+at

Where

t = time

t=\dfrac{v-u}{a}

t=\dfrac{27\ m/s-22\ m/s}{5.93\ m/s^2}

t = 0.84 seconds

Hence, the time required for an object is 0.84 seconds.

4 0
3 years ago
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