Zinc is added to a breaker containing hydrochloric acid and the beaker gets warm what type os reaction is this

Answer:

Dear Kaleb

Answer to your query is provided below

Acceleration of the vehicle is 12m/s^2

Explanation:

Explanation for the same is attached in image

Answer:

Silver.

Explanation:

To determine the identity of the metal, we need to calculate the density of the metal. This is illustrated below:

Mass of metal (m) = 18.15g

Length (L)= 1.2cm

Volume (V) = L³ = 1.2³ = 1.728cm³

Density =.?

The density of a substance is simply defined as the mass of the substance per unit volume of the substance. Mathematically, it is expressed as:

Density = Mass /volume

With the above formula, we can obtain the density of the metal as follow:

Mass = 18.15g

Volume = 1.728cm³

Density =.?

Density = Mass /volume

Density = 18.15g/1.728cm³

Density of the metal = 10.50g/cm³

Comparing the density of metal obtained with the densities given in the table above, we can see that the density of the metal is the same with that of silver.

Therefore, the metal is silver.

Answer:

0.19rad/s and Yes

Explanation:

From the principle of conservation of momentum it means momentum before and after collision is the same.

Momentum before collision is 0.700 kg×12 = 8.4Ns

Momentum of the door = mass of door × velocity of door

8.4Ns = mass of door × velocity of door

Velocity of door = 8.4Ns/45 =0.19m/s

But velocity V= w×r ;

w-angular velocity

r- raduis = width

w= 0.19/1m = 0.19rad/s

2. Yes it did because it resisted The moment of inertia and ensued the locking of the door.

Answer:

rt

Explanation:

Answer:

1A,2D,3B

Explanation:

hope this helps

Answer:

v = 19.2 m/s

Explanation:

In order to find the speed of the string you use the following formula:

[tex]f=\frac{v}{2L}[/tex]          (1)

f: frequency of the string = 80.0Hz

v: speed of the wave = ?

L: length of the string = 12.0cm = 0.12m

The length of the string coincides with the wavelength of the wave for the fundamental mode.

Then, you solve for v in the equation (1), and replace the values of the other parameters:

[tex]v=2Lf=2(0.12m)(80.0Hz)=19.2\frac{m}{s}[/tex]

The speed of the wave is 19.2 m/s

Answer:

W = 414 J, correct is B

Explanation:

Work is defined by

        W = ∫ F .dx

where F is the force, x is the displacement and the point represents the dot product

this expression can also be written with the explicit scalar product

        W = ∫ F dx cos θ

where is the angle between force and displacement

for this case as the force is constant

         W = F x cos θ

calculate

         W = 25.0 18.0 cos (-23)

         W = 414 J

the correct answer is B

Answer:

P1 = 0 gage

P2 = 87.9 lb/ft³

Explanation:

Given data

Airplane flying = 200 mph = 293.33 ft/s

altitude height = 5000-ft

air velocity relative to the airplane = 273 mph = 400.4 ft/s

Solution

we know density at height 5000-ft is 2.04 × [tex]10^{-3}[/tex] slug/ft³

so here P1 + [tex]\frac{\rho v1^2}{2}[/tex]  = P2 + [tex]\frac{\rho v2^2}{2}[/tex]

and here

P1 = 0 gage

because P1 = atmospheric pressure

and so here put here value and we get

P1 + [tex]\frac{\rho v1^2}{2}[/tex]  = P2 + [tex]\frac{\rho v2^2}{2}[/tex]

0 + [tex]\frac{2.048 \times 10^{-3} \times 293.33^2}{2}[/tex]  [tex]= P2 + \frac{2.048 \times 10^{-3} \times 400.4^2}{2}[/tex]  

solve it we get

P2 = 87.9 lb/ft³

Answer:

a) [tex]E = 6.024\,USD[/tex], For m kilograms, it is 4184m J., 3600000 joules, b) [tex]i = 88.200\,A[/tex]

Explanation:

a) The amount of heat needed to warm water is given by the following expression:

[tex]Q_{needed} = m_{w}\cdot c_{w}\cdot (T_{f}-T_{i})[/tex]

Where:

[tex]m_{w}[/tex] - Mass of water, measured in kilograms.

[tex]c_{w}[/tex] - Specific heat of water, measured in [tex]\frac{J}{kg\cdot ^{\circ}C}[/tex].

[tex]T_{f}[/tex], [tex]T_{i}[/tex] - Initial and final temperatures, measured in [tex]^{\circ}C[/tex].

Then,

[tex]Q_{needed} = (1440\,kg)\cdot \left(4184\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (40^{\circ}C - 10^{\circ}C)[/tex]

[tex]Q_{needed} = 180748800\,J[/tex]

The energy needed in kilowatt-hours is:

[tex]Q_{needed} = 180748800\,J\times \left(\frac{1}{3600000}\,\frac{kWh}{J} \right)[/tex]

[tex]Q_{needed} = 50.208\,kWh[/tex]

The electric energy required to heat up the water is:

[tex]E = \frac{50.208\,kWh}{0.75}[/tex]

[tex]E = 66.944\,kWh[/tex]

Lastly, the cost of heating a hot tub is: (USD - US dollars)

[tex]E = (66.944\,kWh)\cdot \left(0.09\,\frac{USD}{kWh} \right)[/tex]

[tex]E = 6.024\,USD[/tex]

The heat needed to raise the temperature a degree of a kilogram of water is 4184 J. For m kilograms, it is 4184m J. Besides, a kilowatt-hour is equal to 3600000 joules.

b) The current required for the electric heater is:

[tex]i = \frac{Q_{needed}}{\eta \cdot \Delta V \cdot \Delta t}[/tex]

[tex]i = \frac{180748800\,J}{0.75\cdot (220\,V)\cdot (3.45\,h)\cdot \left(3600\,\frac{s}{h} \right)}[/tex]

[tex]i = 88.200\,A[/tex]

Answer:

e. 400 Hz

Explanation:

In closed organ pipe,  only odd harmonics of fundamental note is possible .

The fundamental frequency is 200 Hz . Then other overtones will be having following frequencies .

200 x 3 , 200 x 5 , 200 x 7 , 200 x 9 etc

600 Hz , 1000 Hz , 1400 Hz  , 1800 Hz .

Frequency not possible is 400 Hz .

Answer:

T = 676 N

Explanation:

Given that: f = 65 Hz, L = 2.0 m, and ρ = 5.0 g[tex]/m^{2}[/tex] = 0.005 kg

A stationary wave that is set up in the string has a frequency of;

f = [tex]\frac{1}{2L}[/tex][tex]\sqrt{\frac{T}{M} }[/tex]

⇒      T = 4[tex]L^{2}[/tex][tex]f^{2}[/tex]M

Where: t is the tension in the wire, L is the length of the wire, f is the frequency of the waves produced by the wire and M is the mass per unit length of the wire.

But M = L × ρ = (2 × 0.005) = 0.01 kg/m

T = 4 × [tex]2^{2}[/tex] ×[tex]65^{2}[/tex] × 0.01

   = 4 × 4 ×4225 × 0.01

   = 676 N

Tension of the wire is 676 N.

Answer:

t = 5.03 s

Explanation:

To find the time interval when Sheila encounter her brother, you first calculate the angular speed of both Sheila and her brother.

You use the following formula:

[tex]\omega = \frac{v}{r}[/tex]

w: angular speed

v: tangential speed

r: radius of the trajectory = 8 m

For  you have:

[tex]\omega=\frac{4m/s}{8m}=0.5\frac{rad}{s}[/tex]

For her brother:

[tex]\omega'=\frac{6m/s}{8m}=0.75\frac{rad}{s}[/tex]

Next, they will encounter to each other when the angular distance of the Brother of sheila equals the angular distance of Sheila in the opposite direction. This can be written as follow:

[tex]\theta=\omega t\\\\\theta'=\omega ' t[/tex]

They encounter for θ = 2π-θ':

[tex]\omega t=2\pi-\omega' t[/tex]

You replace the values of the parameters in the previous equation and solve for t:

[tex]0.5t=2\pi-0.75t\\\\1.25t=2\pi\\\\t=5.026\approx5.03[/tex]

Hence, Sheila encounter her brother in 5.03 s

Answer:

a

The total initial momentum of the two-block system is  [tex]p_t = 67.5 \ kg \cdot m/s^2[/tex]

b

The magnitude of the final velocity of the two-block system [tex]v_f = 1.9014 \ m/s[/tex]

c

 the change ΔK=Kfinal−Kinitial in the two-block system's kinetic energy due to the collision is  

    [tex]\Delta KE =- 847.08 \ J[/tex]

Explanation:

From the question we are told that

    The mass of first  block  is [tex]m_1 = 2.50 \ kg[/tex]

      The initial velocity of first   block is [tex]u_1 = 27.0 \ m/s[/tex]

          The mass of second block is  [tex]m_2 = 33.0\ kg[/tex]

          initial velocity of second block is  [tex]u_2 = 0 \ m/s[/tex]

The magnitude of the of the total initial momentum of the two-block system is mathematically repented as

        [tex]p_i = (m_1 * u_1 ) + (m_2 * u_2)[/tex]

substituting values

        [tex]p_i = (2.50* 27 ) + (33 * 0)[/tex]

        [tex]p_t = 67.5 \ kg \cdot m/s^2[/tex]

According to the law of linear momentum conservation

        [tex]p_i = p_f[/tex]

Where  [tex]p_f[/tex] is the total final momentum of the system which is mathematically represented as

       [tex]p_f = (m_+m_2) * v_f[/tex]

Where [tex]v_f[/tex] is the final velocity of the system

      [tex]p_i = (m_1 +m_2 ) v_f[/tex]

substituting values

       [tex]67.5 = (2.50+33 ) v_f[/tex]

        [tex]v_f = 1.9014 \ m/s[/tex]

The change in kinetic energy is mathematically represented as

     [tex]\Delta KE = KE_f -KE_i[/tex]

Where [tex]KE_f[/tex] is the final kinetic energy of the two-body system  which is mathematically represented as

        [tex]KE_f = \frac{1}{2} (m_1 +m_2) * v_f^2[/tex]

substituting values

        [tex]KE_f = \frac{1}{2} (2.50 +33) * (1.9014)^2[/tex]

        [tex]KE_f =64.17 J[/tex]

While [tex]KE_i[/tex] is the initial kinetic energy of the two-body system

     [tex]KE_i = \frac{1}{2} * m_1 * u_1^2[/tex]

substituting values

       [tex]KE_i = \frac{1}{2} * 2.5 * 27^2[/tex]

        [tex]KE_i = 911.25 \ J[/tex]

So

    [tex]\Delta KE = 64.17 -911.25[/tex]

  [tex]\Delta KE =- 847.08 \ J[/tex]

Answer:

= 0.0532 cm^3

Explanation:

The computation of volume of the sphere is shown below:-

[tex]Density = \frac{Mass}{Volume}[/tex]

Where,

Density = 13.6 g/cm^3

Mass of sphere = 0.723 g

now we will put the values into the above formula to reach volume of the sphere which is here below:-

[tex]Volume = \frac{0.723}{13.6}[/tex]

= 0.0532 cm^3

Therefore for computing the volume of the sphere we simply applied the above formula.

From Ohm's law . . . Resistance = (voltage) / (current)

Resistance = (120 volts) / (7.6 Amperes)

Resistance = 15.8 Ω

Answer:

100 J

Explanation:

Work = change in energy

W = ΔKE

W = ½ mv² − ½ mv₀²

W = ½ m (v² − v₀²)

W = ½ (10 kg) ((6 m/s)² − (4 m/s)²)

W = 100 J

Answer:

About 2104m/s

Explanation:

[tex]F=ma \\\\F=\dfrac{2660kg}{2}\cdot 1614m/s=2,146,620N \\\\2,146,620N=1,269kg\cdot a \\\\a\approx 1691m/s \\\\v_f=v_o+at=413m/s+1691m/s=2104m/s[/tex]

Hope this helps!

Answer:

U = 3.806 J

Explanation:

The potential energy between the two charges q1 and q2, is given by the following formula:

[tex]U=k\frac{q_1q_2}{r}[/tex]         (1)

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

q1 = 5.1*10^-6 C

q2 = 2.51*10^-6 C

r: distance of separation between particles = 3.02cm = 3.02*10^-2 m

You replace the values of all parameters in the equation (1):

[tex]U=(8.98*10^9Nm^2/C^2)\frac{(5.1*10^{-6}C)(2.51*10^{-6}C)}{3.02*10^{-2}m}\\\\U=3.806J[/tex]

The potential energy of the two particle system is 3.806 J

Answer:

The value of m is 2635.294 grams.

Explanation:

Let suppose that mass-spring system has a simple harmonic motion, to this respect the formula for frequency is:

[tex]f = \frac{\omega}{2\pi}[/tex]

Where [tex]\omega[/tex] is the angular frequency, measured in radians per second.

For a mass-spring system under simple harmonic motion, the angular frequency is:

[tex]\omega = \sqrt{\frac{k}{m} }[/tex]

Where:

[tex]k[/tex] - Spring constant, measured in newtons per meter.

[tex]m[/tex] - Mass, measured in kilograms.

The following equation is obtained after replacing angular frequency in frequency formula:

[tex]f = \frac{1}{2\pi}\cdot \sqrt{\frac{k}{m} }[/tex]

As this shows, frequency is inversely proportional to the square root of mass. Hence, the following relationship is deducted:

[tex]f_{1}\cdot \sqrt{m_{1}} = f_{2} \cdot \sqrt{m_{2}}[/tex]

If [tex]m_{2} = m_{1} + 700\,g[/tex], [tex]f_{1} = 0.72\,hz[/tex] and [tex]f_{2} = 0.64\,hz[/tex], the resulting expression is simplified and then initial mass is found after clearing it:

[tex]f_{1} \cdot \sqrt{m_{1}} = f_{2} \cdot \sqrt{m_{1}+700\,g}[/tex]

[tex]f_{1}^{2} \cdot m_{1} = f_{2}^{2}\cdot (m_{1} + 700\,g)[/tex]

[tex]\left(\frac{f_{1}}{f_{2}} \right)^{2}\cdot m_{1} = m_{1} + 700\,g[/tex]

[tex]\left[\left(\frac{f_{1}}{f_{2}}\right)^{2} - 1\right]\cdot m_{1} = 700\,g[/tex]

[tex]m_{1} = \frac{700\,g}{\left(\frac{f_{1}}{f_{2}} \right)^{2}-1}[/tex]

[tex]m_{1} = \frac{700\,g}{\left(\frac{0.72\,hz}{0.64\,hz} \right)^{2}-1}[/tex]

[tex]m_{1} = 2635.294\,g[/tex]

The value of m is 2635.294 grams.

Answer:

B

Explanation:

Because the force has 2 components (horizontal and vertical), the horizontal component must be smaller than the total force. The Pythagorean theorem only adds positive values (because they're squared) so it makes sense. Using trigonometry, 100*cos(-20) yields a horizontal force of around 939.7N, which is less than 1000N.

Answer:

The terminal velocity is [tex]v_t =17.5 \ m/s[/tex]

Explanation:

From the question we are told that

       The mass of the squirrel is  [tex]m_s = 50\ g = \frac{50}{1000} = 0.05 \ kg[/tex]

      The surface area is   [tex]A_s = 935 cm^2 = \frac{935}{10000} = 0.0935 \ m^2[/tex]

       The height of fall is  h =4.8 m

        The length of the prism is [tex]l = 23.2 = 0.232 \ m[/tex]

          The width of the prism is [tex]w = 11.6 = 0.116 \ m[/tex]

The terminal velocity is mathematically represented as

       [tex]v_t = \sqrt{\frac{2 * m_s * g }{\dho_s * C * A } }[/tex]

Where [tex]\rho[/tex]  is the density of a rectangular prism with a constant values of [tex]\rho = 1.21 \ kg/m^3[/tex]

            [tex]C[/tex] is the drag coefficient for a horizontal skydiver with a value = 1

            A  is the area of the prism the squirrel is assumed to be which is mathematically represented as

      [tex]A = 0.116 * 0.232[/tex]

       [tex]A = 0.026912 \ m^2[/tex]

 substituting values

      [tex]v_t = \sqrt{\frac{2 * 0.510 * 9.8 }{1.21 * 1 * 0.026912 } }[/tex]

     [tex]v_t =17.5 \ m/s[/tex]

the answer is 1.5 hope this helps

Answer:

1.5

Explanation:

0.5=σ/(15/0.5)

σ=3/2 or 1.5

Answer:

t = 25.5 min

Explanation:

To know how many minutes does Richard save, you first calculate the time that Richard takes with both velocities v1 = 65mph and v2 = 80mph.

[tex]t_1=\frac{x}{v_1}=\frac{150mi}{65mph}=2.30h\\\\t_2=\frac{x}{v_2}=\frac{150mi}{80mph}=1.875h[/tex]

Next, you calculate the difference between both times t1 and t2:

[tex]\Delta t=t_1-t_2=2.30h-1.875h=0.425h[/tex]

This is the time that Richard saves when he drives with a speed of 80mph. Finally, you convert the result to minutes:

[tex]0.425h*\frac{60min}{1h}=25.5min=25\ min\ \ 30 s[/tex]

hence, Richard saves 25.5 min (25 min and 30 s) when he drives with a speed of 80mph

Answer:

If you pull a permanent magnet rapidly away from a tank circuit, what is likely to happen in that circuit?

Charge will oscillate in the tank's capacitor and inductor.

Explanation:

Answer:

A) The two potatoes cover the same horizontal distance from the launcher.

B) Mark's potato spends more time in the air than Marvin's potato before hitting the ground.

C) Marvin's potato hits the ground with a greater speed than Mark's potato

Explanation:

A) For projectile motion, the final horizontal distance of the projectile from where it was initially launched (its range) is given as

R = (u² sin 2θ)/g

where

u = initial velocity of the projectile

θ = angle above the horizontal at which the projectile was launched = 30°, 60°

g = acceleration due to gravity on Mars

Since, u and g are the same for Mark and Marvin, sin 2θ would determine which range is higher.

Sin (2×60°) = sin 120°

Sin (2×30°) = sin 60°

Sin 120° = Sin 60°

Hence, the two potatoes cover the same horizontal distance from the launcher.

B) Time spent in the air for a projectile is given as

T = (2u sin θ)/g

Again, since u and g are the same for Mark and Marvin on Mars, sin θ will give the required idea of whose potato spends more time in the air.

Sin 60° = 0.866

Sin 30° = 0.50

Sin 60° > Sin 30°

Hence, Mark's potato spends more time in the air than Marvin's potato.

C) The horizontal velocity for projectile motion is constant all through the motion and is equal to u cos θ

u cos 60° < u cos 30°

And the initial vertical velocity is u sin θ

Final vertical velocity

= (initial vertical velocity) - gt

g = acceleration due to gravity on Mars

T = time of flight

For Mark,

initial vertical velocity = u sin 60°, greater than Marvin's u sin 30°

And Mark's potato's time of flight is greater as established in (B) above.

But for Marvin

initial vertical velocity = u sin 30°, less than Mark's u sin 60°

And Marvin's potato's time of flight is lesser as established in (B) above

So, at the end of the day, the final vertical velocity is almost the same for both Mark's and Marvin's potatoes.

Hence, the horizontal component of the final velocity edges the final speed of the potatoes just before hitting the ground in Marvin's favour.

Hope this Helps!!!

Answer:

The  angular displacement  is  [tex]\theta = 29.6 \ rad[/tex]

Explanation:

From the question we are told that

     The initial angular speed is  [tex]w = 5.35 \ rad/s[/tex]

      The angular acceleration is  [tex]\alpha = 0.331 rad /s^2[/tex]

      The time take is  [tex]t = 4.81 \ s[/tex]

Generally the angular displacement is mathematically represented as

          [tex]\theta = w * t + \frac{1}{2} \alpha * t^2[/tex]

substituting values

         [tex]\theta = 5.35 * 4.81 + \frac{1}{2} * 0.331 * (4.81)^2[/tex]

         [tex]\theta = 29.6 \ rad[/tex]

Answer:

The length of organ pipe A is [tex]L = 0.3611 \ m[/tex]

The length of organ pipe B is  [tex]L_b = 0.2708 \ m[/tex]

Explanation:

From the question we are told that

    The fundamental frequency is  [tex]f = 475 Hz[/tex]

     The speed of sound is  [tex]v_s = 343 \ m/s[/tex]

The fundamental frequency of the organ pipe A  is mathematically represented as

        [tex]f= \frac{v_s}{2 L}[/tex]

Where L is the length of  organ pipe

   Now  making L the subject

        [tex]L = \frac{v_s}{2f}[/tex]

substituting values

        [tex]L = \frac{343}{2 *475}[/tex]

        [tex]L = 0.3611 \ m[/tex]

The second harmonic frequency of the  organ pipe A is mathematically represented as

       [tex]f_2 = \frac{v_2}{L}[/tex]

The third harmonic frequency of the  organ pipe B is mathematically represented as      

      [tex]f_3 = \frac{3 v_s}{4 L_b }[/tex]

So from the question

       [tex]f_2 = f_3[/tex]

So

    [tex]\frac{v_2}{L} = \frac{3 v_s}{4 L_b }[/tex]

Making  [tex]L_b[/tex] the subject

     [tex]L_b = \frac{3}{4} L[/tex]

substituting values

    [tex]L_b = \frac{3}{4} (0.3611)[/tex]

    [tex]L_b = 0.2708 \ m[/tex]

Answer:

Air resistance

Answer B is correct

Explanation:

The friction that occurs when air pushes against a moving object causing it to negatively accelerate is called as air resistance.

hope this helps

brainliest appreciated

good luck! have a nice day!

Answer:

vBxf = 0.08625m/s

Explanation:

This is a problem about the momentum conservation law. The total momentum before equals the total momentum after.

[tex]p_f=p_i[/tex]

pf: final momentum

pi: initial momentum

The analysis of the momentum conservation is about a horizontal momentum (x axis). When the quarterback jumps straight up, his horizontal momentum is zero. Then, after the quarterback throw the ball the sum of the momentum of both quarterback and ball must be zero.

Then, you have:

[tex]m_Qv_{Qxi}+m_{Bxi}v_{Bxi}=m_Qv_{Qxf}+m_{Bxf}v_{Bxf}[/tex]    (1)

mQ: the mass of the quarterback = 80kg

mB: the mass of the football = 0.43kg

(vQx)i: the horizontal velocity of quarterback before throwing the ball = 0m/s

(vBx)i: the horizontal velocity of football before being thrown = 0m/s

(vQx)f: the horizontal velocity of quarterback after throwing the ball = ?

(vBx)f: the horizontal velocity of football after being thrown = 15 m/s

You replace the values of the variables in the equation (1), and you solve for (vBx)f:

[tex]0\ kgm/s=-(80kg)(v_{Bxf})+(0.46kg)(15m/s)\\\\v_{Bxf}=\frac{(0.46kg)(15m/s)}{80kg}=0.08625\frac{m}{s}[/tex]

Where you have taken the speed of the quarterback as negative because is in the negative direction of the x axis.

Hence, the speed of the quarterback after he throws the ball is 0.08625m/s