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Leona [35]
3 years ago
10

A stunt man jumps from the top of a building and lands 10 meters below his initial height. In case A, the stunt man lands on a s

tack of foam mattresses. In case B, the stunt man lands on a stack of concrete blocks. In which case will the person be more likely to sustain serious injury?
Physics
1 answer:
krek1111 [17]3 years ago
8 0

Answer:

Explanation:

The stunt will likely sustain serious injury in case of concrete blocks because the average force acting on the person will be more because concrete blocks do not squeeze to provide more time for the force to act on the body instead it acts for a small amount of interval.

Impulse=F_{avg}\times \Delta T

As impulse is constant so time requires to act force on the body is more as compared to concrete block and thus average force in mattress case is less.  

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If the absolute pressure of a gas is 550.280 kPa, its gage pressure is
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If the absolute pressure of a gas is 550.280 kPa, its gage pressure is <span> a. 101.325 kPa.
b. 448.955 kPa.
c. 651.605 kPa.
d. 277.280 kPa.</span>

The answer is B.
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Which type of galaxy has the most active star formation?
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3 years ago
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Two solenoids A and B, spaced close to each other and sharing the same cylindrical axis, have 430 and 610 turns, respectively. A
KIM [24]

Answer

given,

Two solenoids A and B

Number of turn

Na = 430 turns          Nb = 610 turns

Current = 2.80 A

Average flux through  A  = 300 μWb

Average of flux through B = 90.0  μ Wb

a) L = \dfrac{N \phi}{I}

   L = \dfrac{610\times 90 \times 10^{-6}}{2.80}

   L =19.6 mH

b) inductance of A

   L = \dfrac{N_A \phi_A}{I_A}

   L = \dfrac{430\times 300 \times 10^{-6}}{2.80}

   L =46 mH

c) magnitude of the emf

    \epsilon_B = -L_B\dfrac{dI}{dT}

    \epsilon_B = -(19.6\times 10^{-3})(0.5)

    \epsilon_B = -9.8\times 10^{-3}\ V

    \epsilon_B = -9.8\ mV

7 0
3 years ago
A body has masses of 0.013kg and 0.012kg in oil and water respectively, if the relative density of oil is 0.875, calculate the m
konstantin123 [22]

Answer:

the mass of the body is 0.02 kg.

Explanation:

Given;

relative density of the oil, \gamma _0 = 0.875

mass of the object in oil, M_o = 0.013 kg

mass of the object in water, M_w = 0.012 kg

let the mass of the object in air = M_a

weight of the oil, W_0 = M_a - 0.013

weight of the water, W_w = M_a - 0.012

The relative density of the oil is given as;

\gamma_0 = \frac{density \ of \ oil }{density \ of \ water} = \frac{W_0}{W_w} = \frac{M_a -0.013}{M_a -0.012} \\\\0.875 = \frac{M_a -0.013}{M_a -0.012}\\\\0.875(M_a - 0.012) = M_a - 0.013\\\\0.875M_a - 0.0105 = M_a -0.013\\\\0.875M_a - M_a = 0.0105 - 0.013\\\\-0.125 M_a = -0.0025\\\\M_a = \frac{0.0025}{0.125} \\\\M_a = 0.02 \ kg

Therefore, the mass of the body is 0.02 kg.

6 0
2 years ago
A truck on the freeway originally moving at 6.6 meters/second accelerates uniformly with acceleration a = 2.8 meters/second2 for
Sonja [21]

Answer:

139.514 metres

Explanation:

Initial velocity of the truck = 6.6 m/s

Acceleration of the truck = 2.8 m/s^2

Time interval = 7.9 s

Therefore we use the formula,

s = ut + 1/2 at^2

*where s(the distance travelled)...u(the initial velocity)...t(the time period)

; s = 6.6(7.9) + 1/2 (2.8)(7.9)^2

; s = 52.14 + 87.374

The distance moved by the truck = 139.514m

8 0
3 years ago
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