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Zigmanuir [339]
3 years ago
10

In February 1955, a paratrooper fell 370 m from an airplane without being able to open his chute but happened to land in snow, s

uffering only minor injuries. Assume that his speed at impact was 56 m/s (terminal speed), that his mass (including gear) was 85 kg, and that the magnitude of the force on him from the snow was at the survivable limit of 1.2×10⁵N.
What are (a) the minimum depth of snow that would have stopped him safely and (b) the magnitude of the impulse on him from the snow?
Physics
1 answer:
Nonamiya [84]3 years ago
7 0

Answer:

(a) 1.11 m

(b) 4760 kgm/s  

Explanation:

height of plane (h) = 370 m

velocity (v) = 56 m/s

mass (m) = 85 kg

force = 1.2 x 10^{5} N = 120,000 N

(a) We can get the minimum depth of snow from the equation below

   force x depth = kinetic energy on impact

 f x d = 0.5 x m x v^{2}

 120000 x d = 0.5 x 85 x 56^{2}

 d= (0.5 x 85 x 56^{2}) ÷ 120000 = 1.11 m

(b) the magnitude of impulse is equal to the momentum of the paratrooper and his gear

 = m x v

= 85 x 56 = 4760 kgm/s  

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Suppose that you have a 680 Ω, a 720 Ω and a 1.20 kΩ resistor. (a) What is the maximum resistance you can obtain by combining th
Delvig [45]

Explanation:

As the given data is as follows.

    R_{1} = 680 \ohm ohm\ohm,    R_{2} = 720 \ohm ohm,

   R_{3} = 1.2 k\ohm = 1200 \ohm   (as 1 k ohm = 1000 m)

(a)   We will calculate the maximum resistance by combining the given resistances as follows.

      Max. Resistance = R_{1} + R_{2} + R_{3}

                                  = (680 + 720 + 1200) \ohm ohm

                                  = 2600 ohm

or,                               = 2.6 k\ohm ohm

Therefore, the maximum resistance you can obtain by combining these is 2.6 k\ohm ohm.

(b)   Now, the minimum resistance is calculated as follows.

      Min. Resistance = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}

                                 = \frac{1}{680} + \frac{1}{720} + \frac{1}{1200}

                                 = 3.683 \times 10^{-3} ohm

Hence, we can conclude that minimum resistance you can obtain by combining these is 3.683 \times 10^{-3} ohm.

3 0
3 years ago
You’re driving down the highway late one night at 20 m/s when a deer steps out onto the road 35 m in front on you. Your reaction
Salsk061 [2.6K]

Answer:

i would juh hit da deer no kap

Explanation:

4 0
4 years ago
A charged object is suspended motionless in the air by the gravitational force pulling it down and an electric force pushing it
Savatey [412]

The charge of the object must be 1.11 \times e^{-5} \text { coulomb }

Answer: Option C

<u>Explanation:</u>

Suppose an electric charge can be represented by the symbol Q. This electric charge generates an electric field; Because Q is the source of the electric field, we call this as source charge. The electric field strength of the source charge can be measured with any other charge anywhere in the area. The test charges used to test the field strength.

Its quantity indicated by the symbol q. In the electric field, q exerts an electric, either attractive or repulsive force. As usual, this force is indicated by the symbol F. The electric field’s magnitude is simply defined as the force per charge (q) on Q.

         Electric field, E=\frac{\text { Force }(F)}{q}

Here, given E = 4500 N/C and F = 0.05 N.

We need to find charge of the object (q)

By substituting the given values, we get

      q=\frac{F}{E}=\frac{0.05 N}{4500 \mathrm{N} / \mathrm{c}}=1.11 \times e^{-5} \text { coulomb }

6 0
3 years ago
Which is not an element? Water , Arsenic, sodium, Aluminum
MrRa [10]

Answer:

water

Explanation:

water is not an element, it is a molecule

3 0
4 years ago
Read 2 more answers
Whats the answer???????????
Leviafan [203]

Answer: Less than 4 ohms

Explanation:

We have three resistors with the following resistance:

R_{1}=4\Omega

R_{2}=6\Omega

R_{3}=8\Omega

Now, when the resistors are connected in parallel, the total resistance R is calculated as follows:

\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}

Isolating R:

R=\frac{R_{1}R_{2}R_{3}}{R_{3}(R_{1}+R_{2})+R_{1}R_{2}}

Rewriting with th known values:

R=\frac{(4\Omega)(6\Omega)(8\Omega)}{8\Omega(4\Omega+6\Omega)+(4\Omega)(6\Omega)}

Finally:

R=1.84 \Omega

Hence, the correct option is less than 4 ohms.

4 0
3 years ago
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