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3 years ago

I WILL REPORT YOU IF YOU ANSWER WITH LINK!!! WILL GIVE BRAINLIEST

Physics

1 answer:

mars1129 [50]3 years ago

8 0

Answer:

I think is is

Explanation:

B and C why because i have a gut feeling

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The items in a mixture can be returned to their original form.<br> True<br> False

Grace [21]

I believe it’s false.

4 0

3 years ago

Read 2 more answers

Flapping flight is very energy intensive. A wind tunnel test

steposvetlana [31]

Answer:

The metabolic power for starting flight=134.8W/kg

Explanation:

We are given that

Mass of starling, m=89 g=89/1000=0.089 kg

1 kg=1000 g

Power, P=12 W

Speed, v=11 m/s

We have to find the metabolic power for starting flight.

We know that

Metabolic power for starting flight= $\frac{P}{m}$

Using the formula

Metabolic power for starting flight= $\frac{12}{0.089}$

Metabolic power for starting flight=134.8W/kg

Hence, the metabolic power for starting flight=134.8W/kg

4 0

3 years ago

Which of these is NOT a possible type of energy transformation?

Orlov [11]

to be franc i really think the answer is B

6 0

3 years ago

Automobiles must be able to sustain a frontal impacl The automobile design must allow low speed impacts with little sustained da

valentinak56 [21]

Answer: the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m

Explanation:

Given that;

mass of vehicle m = 1000 kg

for a low speed test; V = 2.5 m/s

bumper maximum deflection = 4 cm = 0.04 m

First we determine the energy of the vehicle just prior to impact;

W_v = 1/2mv²

we substitute

W_v = 1/2 × 1000 × (2.5)²

W_v = 3125 J

now, the the effective design stiffness k will be:

at the impact point, energy of the vehicle converts to elastic potential energy of the bumper;

hence;

W_v = 1/2kx²

we substitute

3125 = 1/2 × k (0.04)²

3125 = 0.0008k

k = 3125 / 0.0008

k = 3906250 N/m

Therefore, the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m

3 0

3 years ago

A railroad car of mass 2.50 3 104 kg is moving with a speed of 4.00 m/s. It collides and couples with three other coupled railro

ipn [44]

Answer:2.5 m/s

37.5KJ

Explanation:

Let $u_1, u_2 , v_f$ be the initial velocity of rail road car ,coupled cars & Final velocity of system respectively.

$m=2.50\times 10^{4}$

Conserving momentum

$mu_1+3mu_2=4mv_f$

$4m+6m=4mv_f$

$v_f=2.5 m/s$

Therefore Final velocity of system is 2.5m/s

(b)Mechanical Energy lost =Initial Kinetic Energy -Final Kinetic Energy

$Initial Kinetic Energy=\frac{1}{2}m\left ( 4^2\right )+\frac{1}{2}m\left ( 2^2\right )=14m J$

$Final Kinetic Energy=\frac{1}{2}4m\left ( 2.5^2\right )=12.5m J$

$Mechanical Energy lost=14m-12.5m=3.75\times 10^4=37.5 KJ$

4 0

3 years ago