A negatively charged rod is brought near a neutral metal sphere. Which of the following is true? A negatively charged rod is brought near a neutral metal sphere. Which of the following is true? There is no electric force between the rod and sphere. There is a repulsive force between the rod and sphere. There is an attractive force between the rod and sphere.

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"The acceleration of a bus is given by ax(t)=αt, where α = 1.28m/s3 is a constant. Part A If the bus's velocity at time t1 = 1.13s is 5.09m/s , what is its velocity at time t2 = 2.02s ? If the bus's position at time t1 = 1.13s is 5.92m , what is its position at time t2 = 2.02s ?"

A) 6.88 m/s

The velocity of the bus can be found by integrating the acceleration. Therefore:

[tex]v(t) = \int a(t) dt = \int (\alpha t)dt=\frac{1}{2}\alpha t^2+C[/tex] (1)

where

[tex]\alpha = 1.28 m/s^3[/tex]

C is a constant term

We know that at [tex]t_1 = 1.13 s[/tex], the velocity is [tex]v=5.09 m/s[/tex]. Substituting these values into (1), we can find the exact value of C:

[tex]C=v(t) - \frac{1}{2}\alpha t^2 = 5.09 - \frac{1}{2}(1.28)(1.13)^2=4.27 m/s[/tex]

So now we can find the velocity at time [tex]t_2 = 2.02 s[/tex]:

[tex]v(2.02)=\frac{1}{2}(1.28)(2.02)^2+4.27=6.88 m/s[/tex]

B) 11.2 m

To find the position, we need to integrate the velocity:

[tex]x(t) = \int v(t) dt = \int (\frac{1}{2}\alpha t^2 + C) dt = \frac{1}{3}\alpha t^3 + Ct +D[/tex] (2)

where D is another constant term.

We know that at [tex]t_1 = 1.13 s[/tex], the position is [tex]v=5.92 m[/tex]. Substituting these values into (2), we can find the exact value of D:

[tex]D = x(t) - \frac{1}{6}\alpha t^3 -Ct = 5.92-\frac{1}{6}(1.28)(1.13)^3 - (4.27)(1.13)=0.79 m[/tex]

And so now we can find the position at time [tex]t_2 = 2.02 s[/tex] using eq.(2):

[tex]x(2.02)=\frac{1}{6}(1.28)(2.02)^3 + (4.27)(2.02) +0.79=11.2 m[/tex]

The question deals with calculating the time-dependent acceleration of a bus using the equation ax(t)=αt, applicable to Physics (specifically mechanics), and is suitable for a high school level.

The question pertains to the acceleration of an object, specifically a bus, which makes this a Physics problem. Acceleration is defined as the change in velocity over time and is a vector quantity having both magnitude and direction. Given the equation ax(t)=αt, where α is a constant equal to 1.28 m/s3, we can recognize that the acceleration is linearly increasing over time because for each second, the acceleration increases by 1.28 m/s2. This kind of problem typically appears in high school Physics.

For instance, to calculate the acceleration of a bus at a specific time point, we simply multiply the constant α by the time t at which we want to know the acceleration. So if we wanted to know the acceleration at t=15 s, we would calculate it as ax(15s) = 1.28 m/s3 × 15 s = 19.2 m/s2. This demonstrates that as time progresses, the acceleration value increases.

Answer:

75 N

Explanation:

t = Time taken = 2 seconds

u = Initial velocity

v = Final velocity

s = Displacement = 30 m

a = Acceleration

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 30=0\times 2+\frac{1}{2}\times a\times 2^2\\\Rightarrow a=\frac{30\times 2}{2^2}\\\Rightarrow a=15\ m/s^2[/tex]

The acceleration due to gravity on the planet is 15 m/s²

Force

F = ma

[tex]F=5\times 15\\\Rightarrow F=75\ N[/tex]

The gravitational force exerted on the object near the planet’s surface is 75 N

We have that for the Question it can be said that he gravitational force exerted on the 5kg object near the planet’s surface

F=75N

From the question we are told

A 5kg object is released from rest near the surface of a planet such that its gravitational field is considered to be constant. The mass of the planet is unknown. After 2s, the object has fallen 30m.

Air resistance is considered to be negligible. What is the gravitational force exerted on the 5kg object near the planet’s surface?

Generally the equation for the Motion   is mathematically given as

[tex]s=ut+1/2at^2\\\\Therefore\\\\30=0+1/2(a)(2)^2\\\\a=15m/s^2[/tex]

Therefore

F=ma

F=5*15

F=75N

Hence, the gravitational force exerted on the 5kg object near the planet’s surface

F=75N

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Answer:

A.[tex]a=203.14\ \frac{m}{s^2}[/tex]

B.s=397.6 m

Explanation:

Given that

speed  u= 284.4 m/s

time t = 1.4 s

here he want to reduce the velocity from 284.4 m/s to 0 m/s.

So the final speed v= 0 m/s

We know that

v= u + at

So now by putting the values

0 = 284.4 -a x 1.4     (here we take negative sign because this is the case of de acceleration)

[tex]a=203.14\ \frac{m}{s^2}[/tex]

So the acceleration  while stopping will be [tex]a=203.14\ \frac{m}{s^2}[/tex].

Lets take distance travel before come top rest is s

We know that

[tex]v^2=u^2-2as[/tex]

[tex]0=284.4^2-2\times 203.14\times s[/tex]

s=397.6 m

So the distance travel while stopping is 397.6 m.

Answer:

[tex]a) acceleration = -203.14 m/s^2\\\\b) distance = 199.1 m[/tex]

Explanation:

Given

initial speed u = 284.4 m/s

final speed v = 0 m/s

duration t = 1.4 s

Solution

a)

Acceleration

[tex]a = \frac{v - u}{t} \\\\a = \frac{0-284.4}{1.4} \\\\a = -203.14 m/s^2[/tex]

b)

Distance

[tex]v^2  - u^2 = 2as\\\\0^2 - 284.4^2 = 2 \times  (-203.14) \times S\\\\S = 199.1 m[/tex]

Answer:

a) It takes 6,37 s b) The Velocity is -59,43 m/s

Explanation:

The initial variables of the balloon are:

Xo = 0 m

Vo = 3 m/s

After one minute the situation is the following:

t= 60 s

X1 = Xo + Vo*t

X1= 0 m + 3 m/s * 60s

X1= 180m

So when the bag falls, its initial variables are the following:

Xo = 180m

X1 = 0m

Vo = 3 m/s

V1= ?

a= -9,8 m/s2

The ecuation of movement for this situation is:

X = Vo*t + 1/2 a*[tex]t^{2}[/tex]

So:

-180m = 3m/s*t+ 1/2*-9,8 m/s2 * [tex]t^{2}[/tex]

To solve this we have

a=-9,8/2

b=3

c=180

The formula is:[tex](-b +/- \sqrt{b^{2} -4ac}) /2a[/tex]

Replacing, we get to 2 solutions, where only the positive one is valid because we are talking about time.

So the answer a) is t= 6,37 s

With that answer we can find the question b), with the following movement formula.

Vf = Vo + at

Vf = +3 m/s + (-) 9,8 m/s2 *6,37s

b) Vf = -59,43 m/s

(a) The time taken for the sandbag to reach the ground is 6.37 s.

(b) The velocity of the sandbag when it hits the ground is -59.42 m/s.

(a) The initial velocity of the balloon is, [tex]u = 3\,m/s[/tex].

Given that the motion of the balloon is constant, i.e.; [tex]a = 0\,m/s[/tex]

So, the height of the balloon after [tex]t = 1\, min = 60\,s[/tex] can be calculated using the second kinematics equation given by;

[tex]s = ut +\frac{1}{2} at^2[/tex]

Substituting the known values, we get;

[tex]s = 3\,m/s \times 60\,s = 180\,m[/tex]

Now, a sandbag is dropped.

The initial velocity of the sandbag will be 3m/s in the upward direction.

i.e.; [tex]u_s = 3\,m/s[/tex]

The acceleration of the bag is given by the gravitational force of the earth.

[tex]a_s = -g = - 9.8\,m/s^2[/tex]

The second kinematics equation is given by;

[tex]s = ut +\frac{1}{2} at^2[/tex]

Substituting the known values, we get;

[tex]-180= 3t -(\frac{1}{2} \times 9.8\times t^2)\\\\\implies 4.9t^2 -3t -180=0\\\\\implies t = \frac{3\pm\sqrt{9 - (4\times 4.9\times -180)}}{9.8} = 6.37\,s[/tex]

(b) The final velocity of the sandbag can be found using the equation;

[tex]v=u+at[/tex]

Substituting the known values, we get;

[tex]v_s = 3m/s -(9.8m/s^2 \times 6.37\,s)=-59.42\,m/s[/tex]

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Answer:

0.000000113 or 1.13*[tex]10^{-7}[/tex] meters

Explanation:

One nanometer is [tex]10^{-9}[/tex] meters. So 113 nanometers would be 113*[tex]10^{-9}[/tex], or 1.13*[tex]10^{-7}[/tex] meters. That's expessed on "cientific notation." On the "usual" notation, it will be 0.000000113 meters.

Answer:

[tex]v_{i} =28.86\frac{ft}{s}[/tex]

Explanation:

Conceptual analysis

We apply the kinematic formula for an object that moves vertically upwards:

[tex](v_{f} )^{2} =(v_{i} )^{2} -2*g*y[/tex]

Where:

[tex]v_{f}[/tex] : final speed in ft/s

[tex]v_{i}[/tex] : initial speed in ft/s

g: acceleration due to gravity in ft/s²

y: vertical position at any time in ft

Known data

For [tex]v_{f} = 25\frac{ft}{s}[/tex] ,[tex]y=\frac{1}{4} h[/tex]; where h is the maximum height

for y=h,  [tex]v_{f} =0[/tex]

Problem development

We replace  [tex]v_{f} = 25\frac{ft}{s}[/tex]  , [tex]y=\frac{1}{4} h (ft)[/tex] in the formula (1),

[[tex]25^{2} =(v_{i} )^{2} -2*g*\frac{h}{4}[/tex]   Equation (1)

in maximum height(h): [tex]v_{f} =0[/tex], Then we replace in formula (1):

[tex]0=(v_{i} )^{2} - 2*g*h[/tex]

[tex]2*g*h=(v_{i} )^{2}[/tex]

[tex]h=\frac{(v_{i})^{2}  }{2g}[/tex]   Equation(2)

We replace (h) of Equation(2) in the  Equation (1) :

[tex]25^{2} =(v_{i} )^{2} -2g\frac{\frac{(v_{i})^{2}  }{2g} }{4}[/tex]

[tex]25^{2} =(v_{i} )^{2} -\frac{(v_{i})^{2}  }{4}[/tex]

[tex]25^{2} =\frac{3}{4} (v_{i} )^{2}[/tex]

[tex]v_{i} =\sqrt{\frac{25^{2}*4 }{3} }[/tex]

[tex]v_{i} =28.86\frac{ft}{s}[/tex]

Explanation:

Given

launch angle[tex]=32^{\circ}[/tex]

ball lands exactly 224 m

Range of projectile =224 m

[tex]Range =\frac{u^2sin2\theta }{g}[/tex]

[tex]224=\frac{u^2sin64}{9.8}[/tex]

[tex]u^2sin64=224\times 9.8=2195.2[/tex]

[tex]u^2=2442.383[/tex]

u=49.42 m/s

(b)Maximum height reached

[tex]H_{max}=\frac{u^2sin^2\theta }{2g}[/tex]

[tex]H_{max}=\frac{49.42^2\times sin^{2}32}{2\cdot 9.8}[/tex]

[tex]H_{max}=34.99 m\approx 35 m[/tex]

Answer:

Normal force: 167.48 N

Explanation:

        ∑[tex]F_{x}[/tex]:  [tex]F{x}-20 = 0[/tex]  

        ∑[tex]F_{y}[/tex]: [tex]N -W+F_{y}=0[/tex]

       [tex]N = W - Fy[/tex]

        [tex]N = mg -Fsin(\theta)[/tex]

        [tex]Fcos(\theta) -20 = 0[/tex]   ⇒   [tex]Fcos(\theta) = 20[/tex][tex]\theta                  

        [tex]\theta=cos^{-1}(20/F)[/tex]

       [tex]N = 20 x 9.81 - 35sin(55.15)[/tex]  ⇒ [tex]N = 167.48 N[/tex]

it takes approximately [tex]\(1.3863 \times 10^{-4}\)[/tex] seconds for the copper rod to cool to an average temperature of 25°C when exposed to an air stream at 20°C with a heat transfer coefficient of [tex]\(20,000 \, \text{W/m}^2 \cdot \text{K}\)[/tex].

The rate at which the copper rod loses heat can be described by Newton's Law of Cooling, which is given by the formula:

[tex]\[ Q(t) = Q_0 e^{-ht} \][/tex]

where:

[tex]\( Q(t) \)[/tex] is the heat at time[tex]\( t \)\\[/tex],

[tex]\( Q_0 \)[/tex]is the initial heat content,

[tex]\( h \)\\[/tex] is the heat transfer coefficient,

[tex]\( t \)[/tex] is the time.

The heat transfer coefficient [tex]\( h \)[/tex] can be calculated using the formula:

[tex]\[ h = \frac{k}{D} \][/tex]

where:

[tex]\( k \)[/tex] is the thermal conductivity of copper,

[tex]\( D \)[/tex] is the diameter of the rod.

The time ( t ) it takes for the rod to cool to a certain temperature can be determined by solving for ( t) when [tex]\( Q(t) \)[/tex]reaches a specific value. In this case, we want to find ( t ) when the average temperature of the rod is 25°C.

Let's proceed with the calculations:

1. **Calculate [tex]\( h \)[/tex]:**

  The thermal conductivity of copper[tex](\( k \)) is approximately \( 400 \, \text{W/m} \cdot \text{K} \)[/tex].

[tex]\[ h = \frac{k}{D} \][/tex]

 [tex]\[ h = \frac{400 \, \text{W/m} \cdot \text{K}}{0.02 \, \text{m}} \] \[ h = 20,000 \, \text{W/m}^2 \cdot \text{K} \][/tex]

2. **Plug in values and solve for ( t ):**

  The initial temperature of the rod[tex](\( T_0 \))[/tex] is 100°C, the ambient temperature [tex](\( T_{\text{air}} \))[/tex] is 20°C, and the average temperature[tex](\( T \))[/tex] is 25°C. The temperature difference[tex]\( \Delta T \) is \( T_0 - T_{\text{air}} \)[/tex].

[tex]\[ \Delta T = T_0 - T_{\text{air}} = 100°C - 20°C = 80°C \][/tex]

 Now, [tex]\( h \) and \( \Delta T \) are known, and we can rearrange the formula \( Q(t) = Q_0 e^{-ht} \) to solve for \( t \)[/tex]:

[tex]\[ t = -\frac{1}{h} \ln\left(\frac{T - T_{\text{air}}}{T_0 - T_{\text{air}}}\right) \][/tex]

Substituting the values:

[tex]\[ t = -\frac{1}{20,000 \, \text{W/m}^2 \cdot \text{K}} \ln\left(\frac{25°C - 20°C}{100°C - 20°C}\right) \][/tex]

Calculate ( t ) using the natural logarithm function, and you'll get the time it takes for the copper rod to cool to an average temperature of 25°C.

let's calculate the time it takes for the copper rod to cool to an average temperature of 25°C.

[tex]\[ t = -\frac{1}{20,000 \, \text{W/m}^2 \cdot \text{K}} \ln\left(\frac{25°C - 20°C}{100°C - 20°C}\right) \][/tex]

[tex]\[ t = -\frac{1}{20,000 \, \text{W/m}^2 \cdot \text{K}} \ln\left(\frac{5}{80}\right) \][/tex]

Now, calculate the natural logarithm:

[tex]\[ t = -\frac{1}{20,000 \, \text{W/m}^2 \cdot \text{K}} \ln\left(0.0625\right) \][/tex]

[tex]\[ t \approx -\frac{1}{20,000 \, \text{W/m}^2 \cdot \text{K}} \times (-2.7726) \][/tex]

[tex]\[ t \approx \frac{2.7726}{20,000 \, \text{W/m}^2 \cdot \text{K}} \][/tex]

[tex]\[ t \approx 1.3863 \times 10^{-4} \, \text{s} \][/tex]

So, it takes approximately [tex]\(1.3863 \times 10^{-4}\)[/tex] seconds for the copper rod to cool to an average temperature of 25°C when exposed to an air stream at 20°C with a heat transfer coefficient of [tex]\(20,000 \, \text{W/m}^2 \cdot \text{K}\)[/tex].

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The time it would take for the copper rod to cool to an average temperature of 25°C is approximately 22.17 minutes.

To solve this problem, we will use the concept of Newton's law of cooling, which states that the rate of heat loss of a body is directly proportional to the difference in the temperatures between the body and its surroundings. The formula for the heat transfer rate is given by:

[tex]\[ \frac{dQ}{dt} = hA(T_s - T_{\infty}) \][/tex]

where:

[tex]\( \frac{dQ}{dt} \)[/tex] is the rate of heat transfer (W),

h is the heat transfer coefficient [tex](W/m^2\·K)[/tex],

A is the surface area of the object [tex](m^2)[/tex],

[tex]\( T_s \)[/tex] is the surface temperature of the object (°C), and

[tex]\( T_{\infty} \)[/tex] is the temperature of the surrounding air (°C).

The copper rod is a cylinder, so its surface area A can be calculated using the formula for the surface area of a cylinder:

[tex]\[ A = 2\pi rL \][/tex]

where:

r is the radius of the rod, and

L is the length of the rod.

Given that the diameter of the rod is 2.0 cm, the radius r is half of that, which is 1.0 cm or 0.01 m. The length L of the rod is not given, so we will assume it to be L meters.

The initial temperature of the rod is [tex]\( T_{initial} = 100 \°C \)[/tex], and the final temperature we want to reach is [tex]\( T_{final} = 25 \°C \)[/tex]. The temperature difference between the rod and the surrounding air is [tex]\( T_s - T_{\infty} \)[/tex].

The heat lost Q by the rod as it cools from [tex]\( T_{initial} \)[/tex] to [tex]\( T_{final} \)[/tex] can be found using the specific heat capacity of copper (c) and the mass (m) of the rod:

[tex]\[ Q = mc(T_{initial} - T_{final}) \][/tex]

The mass (m) of the rod can be calculated from its density \rho and volume V:

[tex]\[ m = \rho V \][/tex]

[tex]\[ V = \pi r^2 L \][/tex]

The density of copper [tex]\( \rho \)[/tex] is approximately [tex]\( 8930 \) kg/m^3[/tex].

Now, we can equate the heat lost to the heat transfer over time:

[tex]\[ mc(T_{initial} - T_{final}) = hA(T_s - T_{\infty}) \cdot \Delta t \][/tex]

Solving for [tex]\( \Delta t \)[/tex], we get:

[tex]\[ \Delta t = \frac{mc(T_{initial} - T_{final})}{hA(T_s - T_{\infty})} \][/tex]

The specific heat capacity of copper (c) is approximately 385 J/kg·K.

Substituting the expressions for (m) and (A) into the equation for [tex]\( \Delta t \)[/tex], we have:

[tex]\[ \Delta t = \frac{\rho \pi r^2 L c(T_{initial} - T_{final})}{h(2\pi rL)(T_s - T_{\infty})} \][/tex]

Simplifying, we get:

[tex]\[ \Delta t = \frac{\rho r c(T_{initial} - T_{final})}{2h(T_s - T_{\infty})} \][/tex]

Now we can plug in the values:

[tex]\( \rho = 8930 \) kg/m^3[/tex],

r = 0.01 m,

c = 385 J/kg·K,

[tex]T_{initial}[/tex] = 100°C ,

[tex]T_{final}[/tex] = 25°C,

[tex]h = 200 W/m^2\·K[/tex],

[tex]T_{\infty}[/tex] = 20°C.

[tex]\[ \Delta t = \frac{8930 \times 0.01 \times 385 \times (100 - 25)}{2 \times 200 \times (25 - 20)} \][/tex]

[tex]\[ \Delta t = \frac{8930 \times 0.01 \times 385 \times 75}{2 \times 200 \times 5} \][/tex]

[tex]\[ \Delta t = \frac{8930 \times 0.01 \times 385 \times 15}{2 \times 200} \][/tex]

[tex]\[ \Delta t = \frac{8930 \times 0.01 \times 385 \times 15}{400} \][/tex]

[tex]\[ \Delta t = \frac{532095}{400} \][/tex]

[tex]\[ \Delta t = 1330.2375 \text{ seconds} \][/tex]

To express this in minutes, we divide by 60:

[tex]\[ \Delta t = \frac{1330.2375}{60} \text{ minutes} \][/tex]

[tex]\[ \Delta t \approx 22.170625 \text{ minutes} \][/tex]

Explanation:

1. Rotational speed : It is defined as the number of revolutions per unit time i.e. revolution per minute.

2. Tangential speed : Instantaneous linear speed of an object in circular motion.

3. Centripetal force : This force acts on object when it moves in a circle. It is a force that produces uniform circular motion.

4. Centripetal acceleration : center-seeking change in direction of velocity

5. Newton's Second Law : This law defines the magnitude of force acting on the object. It is given by, F = m × a

6. Newton's First Law : An object in motion stays in motion unless an external force acts on it.  

Answer:

1. Rotational speed :.............................  revolutions per minute

2. Tangential speed : Instantaneous linear speed of an object in circular motion.

3. Centripetal force :  ......................force that produces uniform circular motion

4. Centripetal acceleration : center-seeking change in direction of velocity

5. Newton's Second Law :  F = ma

6. Newton's First Law : An object in motion stays in motion unless an external force acts on it.  

Explanation:

1. Rotational speed :.............................................  revolutions per minute

2. Tangential speed : Instantaneous linear speed of an object in circular motion.

3. Centripetal force :  ......................force that produces uniform circular motion

4. Centripetal acceleration : center-seeking change in direction of velocity

5. Newton's Second Law : This law defines the magnitude of force acting on the object. It is given by, F = m × a

6. Newton's First Law : An object in motion stays in motion unless an external force acts on it.  

Answer:

[tex]1.40\times 10^{4}Joules [/tex]

Explanation:

The mechanical energy  is conservated in the absence of non conservative forces. In this case there is an air drag, wich will acount for the lost of energy. In other words:

[tex]\frac{1}{2}mv_{i} ^{2} +mgh_{i}-E_{lost}=\frac{1}{2}mv_{f} ^{2} +mgh_{f}[/tex]

Solving for the Energy lost:

[tex]E_{lost}=\frac{1}{2}mv_{i} ^{2} +mgh_{i}-\frac{1}{2}mv_{f} ^{2} -mgh_{f}[/tex]

And so

[tex]E_{lost}=\frac{1}{2}m(v_{i} ^{2}-v_{f} ^{2}) +mg(h_{i}-h_{f}) [/tex]

substituting with data

[tex]E_{lost}=\frac{1}{2}(64)((27) ^{2}-(25) ^{2}) +(64)(9.8)(17)=1.40\times 10^{4}Joules [/tex]

Hope my answer helps.

Have a nice day!

Answer:

[tex]-0.44 m/s^2[/tex]

Explanation:

First of all, we need to calculate the distance covered by the locomotive during the reaction time, which is

t = 0.46 s

During this time, the locomotive travels at

v = 13 m/s

And the motion is uniform, so the distance covered is

[tex]d_1 = vt = (13)(0.46)=6.0 m[/tex]

The locomotive was initially 200 m from the crossing, so the distance left to stop is now

[tex]d=200 - 6.0 = 194.0 m[/tex]

And now the locomotive has to slow down to a final velocity of [tex]v=0[/tex] in this distance. We can find the minimum deceleration needed by using the suvat equation:

[tex]v^2 - u^2 = 2ad[/tex]

where

v = 0 is the final velocity

u = 13 m/s is the initial velocity

a is the deceleration

d = 194.0 m is the distance to stop

Solving for a,

[tex]a=\frac{v^2-u^2}{2d}=\frac{0^2-13^2}{2(194)}=-0.44 m/s^2[/tex]

Answer:

a) 15.68 m/s

b) 130.34 m

Explanation:

Let's consider the origin of the coordinates at the ground, therefore, the equation of motion of the rock and its velocity are

x(t)= h + vt - (1/2)gt^2

v(t) = v - gt

Where h is the height of the building and v is the initial velocity.

The maximum height is reached when v=0, that is v(1.6s) = 0, and we know that x(7s) = 0

Therefore

0 = v(1.6s) = v - (9.8 m/s^2)(1.6s)

v = 15.68 m/s

and

0 = x(7s) = h + (15.68 m/s )(7 s) - (1/2)(9.8 m/s^2)(49s^2)

h = (1/2)(9.8 m/s^2)(49s^2) -  (15.68 m/s )(7 s)

h = 130.34 m

Final answer:

Using kinematic equations, the initial upward velocity is found to be 15.7 m/s, the maximum height attained above the building is 12.6 m, and the height of the building is calculated to be 172.9 m.

Explanation:

To solve the problem of a rock thrown vertically upward, we can use the physics of kinematic equations for uniformly accelerated motion. We'll assume the acceleration due to gravity (g) is -9.81 m/s², acting downward.

Part (a): Upward Velocity

At the maximum height, the rock's velocity is 0 m/s. The time it takes to reach this point is half of the total time in the air before coming back to the starting height. Therefore, using the equation v = u + g*t (where v is the final velocity, u is the initial velocity, g is acceleration due to gravity, and t is time), we get u = -g*t. Plugging in the values, we find the initial upward velocity (u) to be 15.7 m/s.

Part (b): Maximum Height Above Building

To find the maximum height, use the equation h = ut + (1/2)*g*t². Substituting the values, we get a maximum height of 12.6 m above the building.

Part (c): Height of the Building

The total time the rock is in the air is 7.00 s. We can calculate the total height from which the rock was thrown (the height of the building plus the maximum height reached) using the equation s = ut + (1/2)*g*t² where s stands for displacement. Finally, we find the height of the building to be 172.9 m.

Answer:

350x

Explanation:

In a microscope the objective has higher magnification than the eyepiece so, this is a microscope

The magnification of a microscope is given by the product of the magnifications of the eyepiece and and the objective.

Objective lens magnification = 35x =[tex]m_o[/tex]

Eyepiece magnification = 10x =[tex]m_e[/tex]

Total magnification

[tex]M=m_o\times m_e\\\Rightarrow M=35\times 10\\\Rightarrow M=350[/tex]

Total magnification is 350x

The total magnification of the microscope with a 35X objective lens and a 10X ocular lens is 350X.

The subject of the question is the calculation of total magnification in a compound light microscope. The total magnification of a microscope can be determined by multiplying the magnification of the objective lens by the magnification of the ocular (or eyepiece) lens. For the specified microscope with a 35X objective lens and a 10X ocular lens, the total magnification is simply the product of these two numbers.

To calculate the total magnification, use the following formula:

Total magnification = (Objective lens magnification) x (Ocular lens magnification)

Total magnification = (35X) x (10X)

Total magnification = 350X

Therefore, when using a 35X objective lens and a 10X ocular lens, the total magnification of the system is 350X.

Answer:

Fx = 98 lb

Fy = 69 lb

Explanation:

We can see it in the pic

Answer:

98 lb, 69 lb

Explanation:

Tension in wire, T = 120 lb

Angle made with the horizontal, θ = 35°

The horizontal component is given by

[tex]T_{x}=T Cos35[/tex]

[tex]T_{x}=120 Cos35=98.3 lb[/tex]

[tex]T_{x}=98 lb[/tex]

The vertical component is given by

[tex]T_{y}=T Sin35[/tex]

[tex]T_{y}=120 Sin35=68.8 lb[/tex]

[tex]T_{y}=69 lb[/tex]

Thus, the horizontal component of tension is given by 98 lb and the vertical component of tension is given by 69 lb.

Answer:

89.16pounds

Explanation:

The equation that defines this problem is as follows

W=k/X^2

where

W=Weight

K= proportionality constant

X=distance from the center of Earth

first we must find the constant of proportionality, with the first part of the problem

k=WX^2=131x3960^2=2054289600pounds x miles^2

then we use the equation to calculate the woman's weight with the new distance

W=2054289600/(4800)^2=89.16pounds

Answer:

[tex]\Delta S = 2.11 J/K[/tex]

Explanation:

As we know that entropy change for phase conversion is given as

[tex]\Delta S = \frac{\Delta Q}{T}[/tex]

Here we know that heat required to change the phase of the water is given as

[tex]\Delta Q = mL[/tex]

here we have

[tex]m = 0.349 g = 3.49\times 10^{-4} kg[/tex]

L = 2250000 J/kg

now we have

[tex]\Delta Q = (3.49 \times 10^{-4})\times 2250000[/tex]

[tex]\Delta Q = 788.9 J[/tex]

also we know that temperature is approximately same as boiling temperature

so we have

[tex]T = 373 k[/tex]

so here we have

[tex]\Delta S = \frac{788.9}{373}[/tex]

[tex]\Delta S = 2.11 J/K[/tex]

Answer : The cell emf for this cell is 0.077 V

Solution :

The balanced cell reaction will be,  

Oxidation half reaction (anode):  [tex]Zn(s)\rightarrow Zn^{2+}+2e^-[/tex]

Reduction half reaction (cathode):  [tex]Zn^{2+}+2e^-\rightarrow Zn(s)[/tex]

In this case, the cathode and anode both are same. So, [tex]E^o_{cell}[/tex] is equal to zero.

Now we have to calculate the cell emf.

Using Nernest equation :

[tex]E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}{diluted}}{[Zn^{2+}{concentrated}]}[/tex]

where,

n = number of electrons in oxidation-reduction reaction = 2

[tex]E_{cell}[/tex] = ?

[tex][Zn^{2+}{diluted}][/tex] = 0.0111 M

[tex][Zn^{2+}{concentrated}][/tex] = 4.50 M

Now put all the given values in the above equation, we get:

[tex]E_{cell}=0-\frac{0.0592}{2}\log \frac{0.0111M}{4.50M}[/tex]

[tex]E_{cell}=0.077V[/tex]

Therefore, the cell emf for this cell is 0.077 V

The cell emf, or electromotive force, for a voltaic cell, like the Zn2+-Zn kind described, can be calculated with the Nernst equation. The emf is determined by the standard electrode potential as well as the concentrations of the redox species in each half-cell.

A voltaic cell runs on a spontaneous redox reaction occurring indirectly, with the oxidant and reductant redox couples contained in separate half-cells. In your example, both half-cells are Zn2+-Zn electrodes where the reaction is Zn2+ + 2e− → Zn (s) with a standard electrode potential, E°, of -0.763 V. The cell emf, or electromotive force, is calculated through the Nernst Equation:

Ecell = E° - (RT/nF) * ln(Q)

where Q is the reaction quotient, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred in the redox reaction, and F is Faraday's constant. Since we're dealing with a concentration cell, Q is given by the ratio of the concentrations of the reduced and oxidized species. Depending on the temperature and assuming that Zn2+ is reduced at the cathode and oxidized at the anode, the cell emf can be calculated accordingly.

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Answer:

The plane needs 1,56 seconds to clear the intersection.

Explanation:

This is a case of uniformly accelerated rectilinear motion.

[tex]V_0^{2} = V_f^{2} - 2ad[/tex]

[tex]V_0=\sqrt{V_0^{2} } = ?[/tex]

Vf=50 m/s

[tex]V_f^{2}  = (50 m/s)^{2} = 2500  m^{2}/s^{2}[/tex]

a = -5.4 [tex]m/s^{2}[/tex] (Negative because is decelerating)

d = displacement needed to clear the intersection. It should be the width of the intersection plus the lenght of the plane.

d= 59,7m + 25 m = 84.7 m

Calculating [tex]V_0[/tex]:

[tex]V_0^{2} = V_f^{2} - 2ad[/tex]

[tex]V_0^{2}= 2500 \frac{m^{2} }{s^{2} } - 2(-5.4\frac{m}{s^{2} })(84.7 m)[/tex]

[tex]V_0^{2}= 3,414.76 \frac{m^{2} }{s^{2} }[/tex]

[tex]V_0= \sqrt{3,414.76} = 58.44 \frac{m}{s}[/tex]

Otherwise:

[tex]t = \frac{V_f-V_0}{a} =\frac{50\frac{m}{s} - 58.44\frac{m}{s}  }{-5.4 \frac{m}{s^{2} } } = 1.56 s[/tex]

The time needed for the plane to clear the intersection is approximately 1.88 s.

Given data:

- Length of the plane (L) = 59.7 m

- Width of the intersection (d) = 25.0 m

- Deceleration (a) = 5.4 m/s²

- Final speed (v) = 50 m/s

The plane decelerates through the intersection. The final speed v and the deceleration a are given. The total distance the plane covers while decelerating through the intersection is the sum of the length of the plane and the width of the intersection:

[tex]\[ \text{Total distance} = L + d = 59.7 \, \text{m} + 25.0 \, \text{m} = 84.7 \, \text{m} \][/tex]

Determine the initial speed [tex](v_0)[/tex]

Using the kinematic equation:

[tex]\[ v^2 = v_0^2 + 2a \Delta x \][/tex]

We rearrange to solve for [tex]\( v_0 \)[/tex]:

[tex]\[ v_0^2 = v^2 - 2a \Delta x \][/tex]

[tex]\[ v_0^2 = (50 \, \text{m/s})^2 - 2 \times 5.4 \, \text{m/s}^2 \times 84.7 \, \text{m} \]\[ v_0^2 = 2500 - 2 \times 5.4 \times 84.7 \]\[ v_0^2 = 2500 - 913.56 \]\[ v_0^2 = 1586.44 \]\[ v_0 = \sqrt{1586.44} \]\[ v_0 \approx 39.83 \, \text{m/s} \][/tex]

Calculate the time (t)

Using the kinematic equation:

[tex]\[ v = v_0 + at \][/tex]

Rearrange to solve for t:

[tex]\[ t = \frac{v - v_0}{a} \]\[ t = \frac{50 \, \text{m/s} - 39.83 \, \text{m/s}}{5.4 \, \text{m/s}^2} \]\[ t = \frac{10.17 \, \text{m/s}}{5.4 \, \text{m/s}^2} \]\[ t \approx 1.88 \, \text{s} \][/tex]

Answer:

For cardboard = 29.4 g/cm²

For aluminium = 113.4 g/cm²

For lead = 193.8 g/cm²

Explanation:

Given:  

Density of the cardboard, d₁ = 0.6 g/cm³

Density of the aluminium, d₂ = 2.7 g/cm³

Density of the lead, d₃ = 11.4 g/cm³

Length of the cardboard,  L₁ = 49 cm

Length of the aluminium, L₂ = 42 cm

Length of the lead, L₃ = 17 cm

Now,

The absorber thickness is calculated as:

= Density × Length

therefore,

For cardboard = d₁ × L₁ = 0.6 × 49 = 29.4 g/cm²

For aluminium = d₂ × L₂ = 2.7 × 42 = 113.4 g/cm²

For lead = d₃ × L₃ = 11.4 × 17 = 193.8 g/cm²

The work function of the metal is approximately 3.67 eV.

The work function, or binding energy, of a metal is the minimum energy required to eject an electron from the metal surface. In the photoelectric effect, the maximum kinetic energy of ejected electrons (photoelectrons) is given by the equation KE = hf - BE, where hf is the photon energy and BE is the work function. To find the work function W0 of the metal, we can use Equation 6.16, which states that the threshold wavelength for observing the photoelectric effect is given by λ = (hc)/(W0), where h is Planck's constant, c is the speed of light, and W0 is the work function.

Given that the maximum wavelength that can still eject electrons is 542 nm, we can rearrange the equation to solve for W0:

W0 = (hc)/λ

Plugging in the values, we have:

W0 = (6.626 × 10^-34 J·s)(3 × 10^8 m/s)/(542 × 10^-9 m)

Simplifying the calculation, we find that the work function W0 of the metal is approximately 3.67 eV.

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Final answer:

The maximum wavelength for ejecting electrons from the metal surface is 542 nm. Using the energy equation for photons, we calculate the work function (W0) to be approximately 2.29 electron volts.

Explanation:

The question relates to the photoelectric effect and the calculation of the work function of a metal when electromagnetic radiation of a certain wavelength is incident upon it. The work function (W0) is the minimum energy needed to eject an electron from the metal surface. According to the question, the maximum wavelength that can eject an electron is 542 nm.

To find the work function in electron volts, we can use the equation:

E(photon) = hc / λ

Where:

E(photon) is the energy of the photon

h is Planck's constant (4.135667696 × [tex]10^-^1^2[/tex] eV·s)

c is the speed of light (3 × 108 m/s)

λ is the wavelength of the photon in meters (542 nm = 542 × [tex]10^-^9[/tex] m)

First, convert the energy into electron volts:

E(photon) = (4.135667696 × [tex]10^-^1^2[/tex] eV·s × 3 × 108 m/s) / (542 × [tex]10^-^9[/tex] m)

After calculation, we find that E(photon) is approximately 2.29 eV. For a photon to eject an electron, its energy must be equal to or greater than the work function of the metal. Therefore, the work function of the metal is also 2.29 eV.

Hey!

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Answer:

B. level of statistical support

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Explanation:

When you create an incorrect conclusion within a study typically means that the study had something to do with a hypothesis. Well, if you have a statistical support and you have error in it, then the the lab is incorrect. One little thing can get someone confused.

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Hope This Helped! Good Luck!

Final answer:

To calculate the angle of the second displacement, typically one would use vector addition and trigonometry, but the question, as presented, does not provide sufficient information to determine this angle.

Explanation:

To find the angle of the second displacement for the particle, we need to consider the information given: the first displacement is 11 m at an angle of 82° from the positive x-axis, and the resulting displacement after the second move is 8.7 m at 135° to the positive x-axis, using the counterclockwise direction as positive.

Using vector addition, the resulting displacement's direction can help us determine the second displacement angle. However, without more information about the specifics of the second displacement (like its magnitude), we cannot calculate the exact angle or vector using typical methods such as vector components or the law of cosines.

Given the problem as stated, assuming standard vector addition, a common approach would involve drawing a vector diagram and applying trigonometry; but in this case, we seem to be missing details to perform accurate calculations.

Answer:

x = +7, x = -7

Explanation:

The equation to solve is:

[tex]\left|\frac{x}{7}\right|=1[/tex]

Which means that the absolute value of the fraction [tex]\frac{x}{7}[/tex] must be equal to 1. Since we have an absolute value, we have basically two equations to solve:

1) [tex]\frac{x}{7}=+1[/tex]

Solving this one, we find

[tex]\frac{x}{7}\cdot 7 = 1\cdot 7 \rightarrow x = +7[/tex]

2) [tex]\frac{x}{7}=-1[/tex]

Solving this one, we find

[tex]\frac{x}{7}\cdot 7 = -1\cdot 7 \rightarrow x = -7[/tex]

So the two solutions are +7 and -7.

The pilot first flies 1.5 hours at 675 mi/h resulting in a distance of 1012.5 miles. Then 'turns' right and flies 2 hours at the same speed for a distance of 1350 miles. Using the law of cosines with these as vectors, the pilot is approximately 1340 miles from the start.

This problem involves trigonometry and the concept of vectors to solve. During the initial 1.5 hours of flight, the pilot would cover a distance of 1.5 hours x 675 mi/h = 1012.5 miles. When she makes a course correction and heads 10° to her right, she then flies another distance of 2 hours x 675 mi/h = 1350 miles.

We can use the law of cosines, which in this case is c² = a² + b² - 2abcos(θ), where θ is the angle between vectors a and b. With a=1012.5 miles, b=1350 miles, and θ=10°, the calculation results in c = 1339.73 miles. Therefore, the distance from her initial position is approximately 1340 miles.

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Answer:

209 m

Explanation:

The y-component of a vector is the magnitude times the sine of the angle.

y = 253 sin 55.8°

y = 209

Answer:

The answer to your question is: 209 m

Explanation:

Data

length = 253 m

d = 55.8°

y - component = ?

Formula

To solve this problem we need to use a right triangle a the trigonometric functions (sine, cosine, tangent, etc)

Now we have to choose among the trigonometric functions which one to use that relates the opposite side and the hypotenuse.

And that one is sineФ = os/h

we clear os = sineФ x h

     os = sine 55.8 x253

     os =  209 m

soh, cah, toa

Answer:

The answer to your question is:

a) t = 3.64 s

b) vox = 31.59 m/s

c) vy = 35.71 m/s

d) v = 47.67 m/s

Explanation:

a) To calculate the time, we know that voy = 0 m/s so we this this formula

   h = voy + 1/2(gt²)

   65  = 0 + 1/2(9.81)(t²)

   65 = 4.905t²

   t² = 65/4.905 = 13.25

   t = 3.64 s

b) To calculate vox  we use this formula vox = d/t ; vox is constant

   vox = 115/3.64 = 31.59 m/s

c) To calculate voy we use the formula    vy = voy + gt

but voy = 0

               vy = gt = 9.81 x 3.64 = 35.71 m/s

d) To calculate v  we use the pythagorean theorem

            c2 = a2 + b2

            c2 = 31.59² + 35.71² = 997.92 + 1275.20

            c2 = 2273.12

              c = 47.67 m/s

Final answer:

A person can produce approximately 85 W of useful electric power using a bicycle-powered generator when factoring in the typical 25% body efficiency and an 85% efficient generator.

Explanation:

A typical person can maintain a steady energy expenditure of 400 W on a bicycle. Assuming a typical efficiency for the body and a generator that is 85% efficient, the useful electric power produced can be calculated as follows:

First, consider the efficiency of the human body. If we assume that the body is 25% efficient (which means that only 25% of the energy consumed by the body is converted to mechanical work, while the rest is lost as heat), then the actual mechanical work a person can produce is 25% of their energy expenditure.

Therefore, the mechanical work output would be:

Next, we account for the efficiency of the generator:

To conclude, a person can produce approximately 85 W of useful electric power with a bicycle-powered generator when considering typical human body efficiency and an 85% efficient generator.

Answer:

The moist filter paper is in charge of preventing evaporation and ensuring the proper saturation of the air of the chromatography chamber.

Explanation:

Thin-layer chromatography (TLC) is a chromatography technique used to separate non-volatile mixtures.

A strip of moist filter paper is put into the chromatography chamber so that its bottom touches the solvent and the paper lies on the chamber wall and reaches almost to the top of the container.

The container is closed and left for a few minutes to let the solvent vapors ascend the moist filter paper and saturate the air in the chamber.

The moist filter paper is in charge of preventing evaporation and ensuring the proper saturation of the air of the chamber.

Answer:

0.006 N

Explanation:

First find the mass of the horse which is the only unknown variable.

Then use it with the new data. Working out is in the attachment.