What is the Mr of HCO3

an object is dropped from a height of 25 meters. at what velocity will it hit the ground? a 7.0 m/s b 11 m/s c 22 m/s d 49 m/s e

kipiarov [429]

Assuming that the object starts at rest, we know the following values:

distance = 25m
acceleration = 9.81m/s^2 [down]
initial velocity = 0m/s

we want to find final velocity and we don't know the time it took, so we will use the kinematics equation without time in it:

Velocity final^2 = velocity initial^2 + 2 × acceleration × distance

Filling everythint in, we have:

Vf^2 = 0^2 + (2)(-9.81)(-25)
The reason why the values are negative is because they are going in the negative direction

Vf^2 = 490.5

Take the square root of that

Final velocity = 22.15m/s which is answer c

6 0

3 years ago

A rectangular coil with 50 turns of conducting wire and a total resistance of 10.0 Ω initially lies in the yz-plane at time t =

castortr0y [4]

Answer:

a) 43.20V

b) 2.71W/s

c) 40.25s

d) 7.77Nm

Explanation:

(a) The emf of a rotating coil with N turns is given by:

$emf=NBA\omega sin(\omega t)$

N: turns

B: magnitude of the magnetic field

A: area

w: angular velocity

the emf max is given by:

$emf_{max}=NBA\omega=(50)(1.80T)(0.200m*0.100m)(24.0rad/s)\\\\emf_{max}=43.20V$

(b) the maximum rate of change of the magnetic flux is given by:

$\frac{d\Phi_B}{dt}=\frac{d(A\cdot B)}{dt}=\frac{d}{dt}(ABcos\omega t)=AB\omega sin(\omega t)\\\\\frac{d\Phi_B}{dt}_{max}=(\pi(0.200*0.100))(1.80T)(24.0rad/s)=2.71\frac{W}{s}$

(c) $emf(t=0.050s)=(50)(1.80T)(0.200m*0.100m)(24rad/s)sin(24.0rad/s(0.050s))\\\\emf(t=0.050s)=40.26V$

(d) The torque is given by:

$\tau=NABIsin\theta\\\\NAB\omega=emf_{max}\\\\\tau=\frac{emf_{max}}{\omega}\frac{emf_{max}}{R}\\\\\tau=\frac{(43.20V)^2}{(24.0rad/s)(10.0\Omega)}=7.77Nm$

3 0

2 years ago

The motion of a piston in an auto engine is simple harmonic. If the piston travels back and forth over a distance of 10 cm and t

chubhunter [2.5K]

Answer:

Maximum force will be 29040 N

Explanation:

We have given mass of the piston m = 1.5 kg

Amplitude A = 10 cm = 0.1 m

Angular speed $\omega =4200rpm=\frac{2\times 3.14\times 4200}{60}=440rad/sec$

We know that angular speed is given by $\omega =\sqrt{\frac{k}{m}}$

$440=\sqrt{\frac{k}{1.5}}$

Squaring both side

$193600=\frac{k}{1.5}$

k = 290400 N/m

Now force is given by $F=kA=290400\times 0.1=29040N$

3 0

3 years ago

Which would most likely cause a decrease in the rate of energy production in a fusion nuclear reactor? a drop in the temperature

True [87]

A drop in temperature would most likely result in a decline in the rate of energy production in a fusion nuclear reactor.

<h3>What is nuclear power?</h3>

The utilization of nuclear reactions to generate energy is known as nuclear power.

Nuclear power facilities produce the great bulk of the electricity generated by nuclear fission of uranium and plutonium.

The environment should be hot enough for the deuterium and tritium ions' kinetic energies to be sufficient to break through the Coulomb barrier and fuse together.

Therefore, a decrease in temperature would most likely cause a decrease in the rate of energy production in a fusion nuclear reactor.

To learn more about nuclear power here;

brainly.com/question/9859575

#SPJ1

6 0

1 year ago

Need help ASAP

zhannawk [14.2K]

For number one, the answer is the first dash mark

And for number two, the answer is the 4th dash.

5 0

2 years ago