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vodomira [7]
3 years ago
5

True or False: Chemical energy stored in food cannot be transformed into mechanical energy

Physics
2 answers:
Novosadov [1.4K]3 years ago
7 0
The answer to this question is false. 
nordsb [41]3 years ago
3 0
It's false. it can be transform

help me to promote as brainliest
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Honey bees can acquire a small net charge on the order of 1 pC as they fly through the air and interact with plants. Estimate th
murzikaleks [220]

Answer:

3.35\cdot 10^{-16}N

Explanation:

The force exerted on a charged particle due to a magnetic field is given by:

F=qvB sin \theta

where

q is the charge of the particle

v is the velocity of the particle

B is the magnetic field strength

\theta is the angle between the directions of B and v

In this problem we have:

q=1 pC=1\cdot 10^{-12} C is the charge of one honey bee

v=15 mi/h = 6.7 m/s is the velocity of the bee

B=5.0\cdot 10^{-5} T is the average strength of the Earth's magnetic field

\theta=90^{\circ}, because the bee flies east to west while the Earth's magnetic field is south to north

Substituting into the equation, we find:

F=(1\cdot 10^{-12}C)(6.7 m/s)(5.0\cdot 10^{-5} T)(sin 90^{\circ})=3.35\cdot 10^{-16}N

7 0
3 years ago
600 J of work is done on a system in a process that decreases the thermal energy of the system by 300 J.
Daniel [21]

Answer:

900 J

Explanation:

\Delta U = Change in entropy = -300 J (Decrease)

W = Work done = 600 J

Q = Heat transferred

Change in internal energy is given by

\Delta U=W+Q\\\Rightarrow Q=\Delta U-W\\\Rightarrow Q=-300-600\\\Rightarrow Q=-900\ J

The heat transferred to or from the system as heat is -900 J.

900 J is transferred. As heat is transferred from the system the sign is negative.

6 0
3 years ago
A 0.060 kg ball hits the ground with a speed of -32m/s. The ball is in contact with the ground for 45 milliseconds and the groun
Dominik [7]
Using the kinematic equation V=Vo + at, the initial velocity (Vo) is -32, the acceleration (a) is found by dividing force by mass 55N/.06kg=916.66 m/s/s, and the time is 0.045 sec.

V=Vo+at
V=-32+916.66*.045
V=9.25

The answer is A.
9 0
4 years ago
Read 2 more answers
What type of energy best represents the strength of an ionic bond? ...?
Julli [10]
I'd say potential/kinetic energy.
5 0
3 years ago
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A puck of mass 0.70 kg approaches a second, identical puck that is stationary on frictionless ice. The initial speed of the movi
natali 33 [55]

Answer:

  • v_1  =  \ 5.196 \frac{m}{s}
  • v_2 =  3 \frac{m}{s}

Explanation:

For this problem, we just need to remember conservation of momentum, as there are no external forces in the horizontal direction:

\vec{p}_i = \vec{p}_f

where the suffix i  means initial, and the suffix f means final.

The initial momentum will be:

\vec{p}_i = m_1 \ \vec{v}_{1_i} + m_2 \ \vec{v}_{2_i}

as the second puck is initially at rest:

\vec{v}_{2_i} = 0

Using the unit vector \vec{i} pointing in the original line of motion:

\vec{v}_{1_i} = 6.0 \frac{m}{s} \hat{i}

\vec{p}_i = 0.70 \ kg  \ 6.0 \frac{m}{s} \ \hat{i} + 0.70 \ kg \ 0

\vec{p}_i = 4.2 \ \frac{kg \ m}{s} \ \hat{i}

So:

\vec{p}_i =  4.2 \ \frac{kg \ m}{s} \ \hat{i} = \vec{p}_f

\vec{p}_f =  4.2 \ \frac{kg \ m}{s} \ \hat{i}

Knowing the magnitude and directions relative to the x axis, we can find Cartesian representation of the vectors using the formula

\ \vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )

So, our velocity vectors will be:

\vec{v}_{1_f} = v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ )

\vec{v}_{2_f} = v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )

We got

\vec{p}_f = 0.7 \ kg \ \vec{v}_{1_f} + 0.7 \ kg \ \vec{v}_{2_f}

4.2 \ \frac{kg \ m}{s} \ \hat{i} = 0.7 \ kg \   v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ )  + 0.7 \ kg \ v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )

So, we got the equations:

4.2 \ \frac{kg \ m}{s}  = 0.7 \ kg \   v_1 \  cos(30 \°) + 0.7 \ kg \ v_2 \  cos(-60 \°)

and

0  = 0.7 \ kg \   v_1 \  sin(30 \°) + 0.7 \ kg \ v_2 \  sin(-60 \°).

From the last one, we get:

0  = 0.7 \ kg \  ( v_1 \  sin(30 \°) +  \ v_2 \  sin(-60 \°) )

0  =  v_1 \  sin(30 \°) +  \ v_2 \  sin(-60 \°)

v_1 \  sin(30 \°) = -  \ v_2 \  sin(-60 \°)

v_1  =  \ v_2 \  \frac{sin(60 \°)}{ sin(30 \°) }

and, for the first one:

4.2 \ \frac{kg \ m}{s}  = 0.7 \ kg  \ (  v_1 \  cos(30 \°) + v_2 \  cos(60 \°) )

\frac{4.2 \ \frac{kg \ m}{s}}{ 0.7 \ kg} =    v_1 \  cos(30 \°) + v_2 \  cos(60 \°)

\frac{4.2 \ \frac{kg \ m}{s}}{ 0.7 \ kg} =    v_1 \  cos(30 \°) + v_2 \  cos(60 \°)

6 \ \frac{m}{s} =    (\ v_2 \  \frac{sin(60 \°)}{ sin(30 \°) } ) \  cos(30 \°) + v_2 \  cos(60 \°)

6 \ \frac{m}{s} = v_2     (\   \frac{sin(60 \°)}{ sin(30 \°) } ) \  cos(30 \°) +   cos(60 \°)

6 \ \frac{m}{s} = v_2  * 2

so:

v_2 = 6 \ \frac{m}{s} / 2 = 3 \frac{m}{s}

and

v_1  =  \ 3 \frac{m}{s}  \  \frac{sin(60 \°)}{ sin(30 \°) }

v_1  =  \ 5.196 \frac{m}{s}

3 0
3 years ago
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