You push downward on a box at an angle 25° below the horizontal with a force of 750 N. If the box is on a flat horizontal surface for which the coefficient of static friction with the box is 0.66, what is the mass of the heaviest box you will be able to move?

Answer:

15435 J

Explanation:

Latent heat of fusion, Lf = 334 J/g

Specific heat of ice, ci = 2.1 J / g C

Latent heat of vaporisation, Lv = 2230 J/g

Specific heat of water, cw = 4.18 J / g C

mass, m = 50 g, T = - 5 degree C

There are following steps

(i) - 5 degree C ice converts into 0 degree C ice

H1 = m x ci x ΔT = 50 x 2.1 x 5 = 525 J

(ii) 0 degree C ice converts into 0 degree C water

H2 = m x Lf = 5 x 334 = 1670 J

(iii) 0 degree C water converts into 100 degree C water

H3 = m x cw x ΔT = 5 x 4.18 x 100 = 2090 J

(iv) 100 degree C water converts into 100 degree C steam

H4 = m x Lv = 5 x 2230 = 11150 J

Total heat required

H = H1 + H2 + H3 + H4

H = 525 + 1670 + 2090 + 11150 = 15435 J

Answer:

3.95 m

Explanation:

m = 1 kg, h = 100 m, k = 125 N/m

Let the spring is compressed by y.

Use the conservation of energy

potential energy of the mass is equal to the energy stored in the spring

m x g x h = 1/2 x ky^2

1 x 9.8 x 100 = 0.5 x 125 x y^2

y^2 = 15.68

y = 3.95 m

Answer:

The magnetic flux density is [tex]2.11\times10^{-6}\ T[/tex]

Explanation:

Given that,

Distance = 0.36 m

Current = 3.8 A

We need to calculate the magnetic flux density

Using formula of magnetic field

[tex]B =\dfrac{\mu_{0}I}{2r}[/tex]

Where,

r = radius

I = current

Put the value into the formula

[tex]B =\dfrac{4\pi\times10^{-7}\times3.8}{2\times\pi\times0.36}[/tex]

[tex]B=2.11\times10^{-6}\ T[/tex]

Hence, The magnetic flux density is [tex]2.11\times10^{-6}\ T[/tex]

The Linear speed of the sphere is mathematically given as

v = 4.1 m/s

Question Parameter(s):

A solid sphere of mass 4.0 kg and radius of 0.12 m starts from rest at the top of a ramp inclined 15°

Generally, the equation for the conservation of energy   is mathematically given as

[tex]mgh = \frac{1}{2}mv^2 + \frac{1}{2}Iw^2[/tex]

Therefore

[tex]mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v^2}{R^2})\\\\mgh = \frac{7}{10}mv^2[/tex]

In conclusion, the Speed is

[tex]v^2 = \sqrt{\frac{10}{7}(9.8)(1.2)}[/tex]

v = 4.1 m/s

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The linear speed of the sphere when it reaches the bottom of the ramp is approximately[tex]\( 4.85 \, \text{m/s} \).[/tex]

At the top of the ramp, the potential energy U  of the sphere is given by:

[tex]\[ U = mgh \][/tex]

where m  is the mass of the sphere, g is the acceleration due to gravity (approximately[tex]\( 9.8 \, \text{m/s}^2[/tex], and  h is the height of the ramp.

At the bottom of the ramp, the kinetic energy K of the sphere is given by:

[tex]\[ K = \frac{1}{2}mv^2 \][/tex]

where  v  is the linear speed of the sphere.

Since energy is conserved, we have:

U = K

[tex]\[ mgh = \frac{1}{2}mv^2 \][/tex]

We can solve for  v  by canceling the mass m from both sides and multiplying through by 2:

[tex]\[ 2gh = v^2 \][/tex]

Taking the square root of both sides gives us the linear speed  v :

[tex]\[ v = \sqrt{2gh} \][/tex]

Given that the height  h  is 1.2 m and the acceleration due to gravity  g  is [tex]\( 9.8 \, \text{m/s}^2 \)[/tex], we can plug in these values:

[tex]\[ v = \sqrt{2 \times 9.8 \, \text{m/s}^2 \times 1.2 \, \text{m}} \][/tex]

[tex]\[ v = \sqrt{23.52 \, \text{m}^2/\text{s}^2} \][/tex]

[tex]\[ v \ =4.85 \, \text{m/s} \][/tex]

Answer:

[tex]V_{rex}=75.65m/s[/tex] and [tex]a_{res}=0[/tex] at t=5 secs

Explanation:

We have r =2m

[tex]\therefore \frac{dr}{dt}=0\\\\=>V_{r}=0[/tex]

Similarly

[tex]=>V_{\theta }=\omega r\\\\\omega =\frac{d\theta }{dt}=\frac{d(\pi t)}{dt}=\pi \\\\\therefore V_{\theta }=\pi r=2\pi[/tex]

Similarly

[tex]=>V_{z }=\frac{dz}{dt}\\\\V_{z}=\frac{dsin(24\pi t)}{dt}\\\\V_{z}=24\pi cos(24\pi t)[/tex]

Hence

at t =5s [tex]V_{\theta}=2\pi m/s[/tex]

[tex]V_{z}=24\pi cos(120\pi)[/tex]

[tex]V_{z}=24\pi m/s[/tex]

[tex]V_{res}=\sqrt{V_{\theta }^{2}+V_{z}^{2}}[/tex]

Applying values we get

[tex]V_{res}=75.65m/s[/tex]

Similarly

[tex]a_{\theta }=\frac{dV_{\theta }}{dt}=\frac{d(2\pi) }{dt}=0\\\\a_{z}=\frac{d^{2}(sin(24\pi t))}{dt^{2}}\\\\a_{z}=-24^{2}\pi^{2}sin(24\pi t)\\\\\therefore t=5\\a_{z}=0[/tex]

Explanation:

Mass of the crate, m = 68 kg

We need to find the resulting acceleration if :

(a) Force, P = 0

P = m a

⇒ a = 0

(b) P = 181 N

[tex]a=\dfrac{P}{m}[/tex]

[tex]a=\dfrac{181\ N}{68\ kg}[/tex]

[tex]a=2.67\ m/s^2[/tex]

(c) P = 352 N

[tex]a=\dfrac{P}{m}[/tex]

[tex]a=\dfrac{352\ N}{68\ kg}[/tex]

[tex]a=5.17\ m/s^2[/tex]

Hence, this is the required solution.

Answer:

Part a)

a = 531.7 m/s/s

Part b)

a = 54.25 g

Explanation:

Part a)

Initial speed of the car is given as

[tex]v = 105 km/h[/tex]

now we have

[tex]v = 29.2 m/s[/tex]

now we know that it stops in 0.80 m

now by kinematics we have

[tex]a = \frac{v_f^2 - v_i^2}{2d}[/tex]

so we will have

[tex]a = \frac{0 - 29.2^2}{2(0.80)}[/tex]

[tex]a = 531.7 m/s^2[/tex]

Part b)

in terms of g this is equal to

[tex]a = \frac{531.7}{9.80}[/tex]

[tex]a = 54.25 g[/tex]

Final answer:

The magnitude of the average acceleration of the driver during the collision is approximately -532.09 [tex]m/s^2[/tex], which is about 54.29 g's when expressed in terms of the acceleration due to gravity.

Explanation:

To calculate the magnitude of the average acceleration of the driver during the collision, we can use the following kinematic equation that relates velocity, acceleration, and distance:

[tex]v^2 = u^2 + 2a * s[/tex]

Where:
v is the final velocity (0 m/s, since the driver comes to a stop)

u is the initial velocity (105 km/h, which needs to be converted to m/s)

a is the acceleration (the quantity we want to find)

s is the stopping distance (0.80 m)

First, convert the velocity from km/h to m/s by multiplying by (1000 m/1 km)*(1 h/3600 s) to get approximately 29.17 m/s. Now we can solve for 'a' as follows:

[tex](0)^2 = (29.17 m/s)^2 + 2 * a * (0.80 m)-29.17^2 = 2 * a * 0.80a = -(29.17)^2 / (2 * 0.80)a = -532.09[/tex]

We find that the magnitude of the average acceleration is approximately [tex]-532.09 m/s^2[/tex]. To express this in terms of 'g's, we divide by the acceleration due to gravity [tex](9.80 m/s^2)[/tex]:

[tex]a_g = -532.09 / 9.80a_g =54.29 g's[/tex]

Answer:

[tex]I = 1.6 kg m^2[/tex]

Explanation:

Moment of inertia of disc is given as

[tex]I = \frac{1}{2}mR^2[/tex]

now we have

m = 20 kg

R = 20.0 cm = 0.20 m

now we have

[tex]I_{disc} = \frac{1}{2}(20 kg)(0.20 m)^2[/tex]

[tex]I_{disc} = 0.4 kg m^2[/tex]

Now the additional mass of 20 kg is placed on its rim so it will behave as a ring so moment of inertia of that part of the disc is

[tex]I = mR^2[/tex]

m = 20 kg

R = 20 cm = 0.20 m

[tex]I_{ring} = 20(0.20^2)[/tex]

[tex]I_{ring} = 0.8 kg m^2[/tex]

Now four point masses each of the mass of one fourth of mass of disc is placed at four positions so moment of inertia of these four masses is given as

[tex]I_{mass} = 4( m'r^2)[/tex]

here we have

[tex]m' = \frac{m}{4}[/tex]

[tex]I_{mass} = 4(\frac{m}{4})(0.10^2 + 0.10^2)[/tex]

[tex]I_{mass} = 20(0.02) = 0.40 kg m^2[/tex]

Now total moment of inertia of the system is given as

[tex]I = I_{disc} + I_{ring} + I_{mass}[/tex]

[tex]I = 0.4 + 0.8 + 0.4 = 1.6 kg m^2[/tex]

Answer:46.03 KJ

Explanation:

no of moles(n)=5

temperature of air(T)=[tex]20^{\circ}[/tex]

pressure(p)=1 atm

final volume is [tex]\frac{V}{10}[/tex]

We know work done in adaibatic process is given by

W=[tex]\frac{P_iV_i-P_fV_f}{\gamma -1}[/tex]

[tex]\gamma[/tex] for air is 1.4

we know [tex]P_iV_i^{\gamma }= P_fV_f^{\gamma }[/tex]

[tex]1\left ( V\right )^{\gamma }[/tex]=[tex]P_f\left (\frac{v}{10}\right )^{\gamma }[/tex]

[tex]10^{\gamma }=P_f[/tex]

[tex]P_f=25.118 atm[/tex]

W=[tex]\frac{1\times 0.1218-25.118\times 0.01218}{1.4 -1}[/tex]

W=-46.03 KJ

it means work is done on the system

Answer:

9.1 x 10⁵ ohm

Explanation:

C = Capacitance of the capacitor = 9 x 10⁻⁶ F  

V₀ = Voltage of the battery = 13 Volts  

V = Potential difference across the battery after time "t" = 4 Volts  

t = time interval = 3 sec  

T = Time constant

R = resistance  

Potential difference across the battery after time "t" is given as  

[tex]V = V_{o} (1-e^{\frac{-t}{T}})[/tex]

[tex]4 = 13 (1-e^{\frac{-3}{T}})[/tex]

T = 8.2 sec  

Time constant is given as  

T = RC  

8.2 = (9 x 10⁻⁶) R  

R = 9.1 x 10⁵ ohm

Final answer:

To determine the resistance R, the RC circuit charging equation is used with the given values. By rearranging the equation and solving, the resistance R is found to be approximately 7.97 kΩ.

Explanation:

To find the resistance R in the given circuit, we use the charging equation for a capacitor in an RC circuit:

V(t) = V_0(1 - e^{-t/RC})

Where V(t) is the voltage across the capacitor at time t, V_0 is the initial voltage provided by the battery, R is the resistance, C is the capacitance, and t is the time.

Plugging in the given values:

V(t) = 4.00 V

V_0 = 13.0 V

C = 9.0 µF

t = 3.00 s

We have:

4.00 = 13.0(1 - e^{-3/(9.0×10^{-6}R)})

Now solve for R:

1 - \frac{4.00}{13.0} = e^{-3/(9.0×10^{-6}R))}

Simplifying:

\frac{9.00}{13.00} = e^{-3/(9.0×10^{-6}R))}

Take the natural logarithm of both sides:

ln(\frac{9.00}{13.00}) = -\frac{3}{9×10^{-6}R}

Multiply by -9×10^{-6}R and divide by 3:

R = -\frac{9×10^{-6}ln(\frac{9.00}{13.00})}{3}

R ≈ 7.97 kΩ

Thus, the resistance R is approximately 7.97 kΩ.

Answer:

605 km

Explanation:

Hello

the same units of measure should be used, then

Step 1

convert  42 m/s ⇒   km/h

1 km =1000 m

1 h = 36000 sec

[tex]42 \frac{m}{s}*\frac{1\ km}{1000\ m}=0.042\ \frac{km}{s}\\ 0.042\ \frac{km}{s}\\[/tex]

[tex]0.042\ \frac{km}{s}*\frac{3600\ s}{1\ h} =151.2 \frac{km}{h}\\ \\Velocity =151.2\ \frac{km}{h}[/tex]

Step 2

find kilometers traveled after 4  hours

[tex]V=\frac{s}{t}\\ \\[/tex]

V,velocity

s, distance traveled

t. time

now, isolating s

[tex]V=\frac{s}{t} \\s=V * t\\[/tex]

and replacing

[tex]s=V * t\\s=151.2\frac{km}{h}*4 hours\\ s=604.8 km\\[/tex]

S=604.8 Km

Have a great day

Answer:

(a) 3.13 m/s

(b) 9.9 m/s

(c) 7.73 m/s

Explanation:

u = 0 m/s, g = 9.8 m/s^2

Let v be the velocity of ball as it hit the ground.

(a) h = 0.5 m

Use third equation of motion.

v^2 = u^2 + 2 g h

v^2 = 0 + 2 x 9.8 x 0.5

v^2 = 9.8

v = 3.13 m/s

(b) h = 5 m

Use third equation of motion.

v^2 = u^2 + 2 g h

v^2 = 0 + 2 x 9.8 x 5

v^2 = 98

v = 9.9 m/s

(c) h = 10 feet = 3.048 m

Use third equation of motion.

v^2 = u^2 + 2 g h

v^2 = 0 + 2 x 9.8 x 3.048

v^2 = 59.74

v = 7.73 m/s

Hey there!:

Take the speed of sound to be 343m/s.

Direct frequency perceived by observer:

(343 + 17) / 343) * 90Hz = 94.460Hz

Change in frequency = ( 4.460 - 90 ) = 4.460Hz.  

90 - (4.460 x 2) = 81.08 Hz. indirect frequency heard by observer.  

Therefore :

Beat frequency = (94.460 - 81.08) = 13.38Hz.  

Hope this helps!

Answer:

Velocity of airplane is 500 km/h

Velocity of wind is 40 km/h

Explanation:

[tex]V_a[/tex]= Velocity of airplane in still air

[tex]V_w[/tex]= Velocity of wind

Time taken by plane to travel 1150 km against the wind is 2.5 hours

[tex]V_a-V_w=\frac {1150}{2.5}\\\Rightarrow V_a-V_w=460\quad (1)[/tex]

Time taken by plane to travel 450 km against the wind is 50 minutes = 50/60 hours

[tex]V_a+V_w=\frac {450}{50}\times 60\\\Rightarrow V_a-V_w=540\quad (2)[/tex]

Subtracting the two equations we get

[tex]V_a-V_w-V_a-V_w=460-540\\\Rightarrow -2V_w=-80\\\Rightarrow V_w=40\ km/h[/tex]

Applying the value of velocity of wind to the first equation

[tex]V_a-40=460\\\Rightarrow V_a =500\ km/h[/tex]

∴ Velocity of airplane in still air is 500 km/h and Velocity of wind is 40 km/h

The given mathematical problem involves calculating speeds. The speed of the airplane in still air was found to be 501 km/h and the wind speed was calculated to be 41 km/h, by using the concept of speed equals distance divided by time.

The subject is related to measurements of speed, which is essentially a mathematical problem, specifically involving algebraic calculations. In order to solve this problem, we will use the concept that speed equals distance divided by time.

Firstly, convert all time measurements to the same unit. Since the speed of an airplane is typically measured in km/h, it will be convenient to convert the 50 minutes to hours (50/60 = 0.83 hours).

When moving against the wind, the combined speed (airplane speed minus wind speed) is 1150 km / 2.5 h = 460 km/h. When moving with the wind, the combined speed (airplane speed plus wind speed) becomes 450 km / 0.83 h = 542 km/h.

Now you can find the speed of the airplane in still air by averaging the two combined speeds: (460 km/h + 542 km/h) / 2 = 501 km/h. The speed of the wind can be found by subtracting the airplane's speed from the higher combined speed: 542 km/h - 501 km/h = 41 km/h. Therefore, the speed of the airplane in still air is 501 km/h and the wind speed is 41 km/h.

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Answer:

Height, h = 13.02 meters

Explanation:

It is given that,

Spring constant of the spring, k = 750 N/m

It is compressed by 35 cm relative to its unstained length, x = 35 cm = 0.35 m

Mass of the object, m = 0.36 kg

When the spring is released, the mass is launched vertically in the air. We need to find the height attained by the mass at this position. On applying the conservation of energy as :

Energy stored in the spring = change on potential energy

[tex]\dfrac{1}{2}kx^2=mgh[/tex]

[tex]h=\dfrac{kx^2}{2mg}[/tex]

[tex]h=\dfrac{750\ N/m\times (0.35\ m)^2}{2\times 0.36\ kg\times 9.8\ m/s^2}[/tex]

h = 13.02 meters

So, the mass will reach a height of 13.02 meters. Hence, this is the required solution.

The maximum height reached by the mass when the spring is released is 13.02 m.

The given parameters;

Apply the principle of conservation energy to determine the maximum height reached by the mass when the spring is released.

mgh = ¹/₂kx²

where;

[tex]h = \frac{kx^2}{2mg} \\\\h = \frac{(750) \times (0.35)^2}{2(0.36\times 9.8)} \\\\h = 13.02 \ m[/tex]

Thus, the maximum height reached by the mass when the spring is released is 13.02 m.

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Answer:

The diameter of a square link is 0.0233 inch.

Explanation:

Given that,

Load = 27000 lbs

Modulus of elasticity [tex]E= 30\times10^{6}\ psi[/tex]

Working stress [tex]\sigma=7000\ psi[/tex]

length l = 55 in

We need to calculate the diameter of a square link

Using formula of stress

[tex] \sigma=\dfrac{Force}{Area}[/tex]

[tex]7000=\dfrac{27000}{\pi\times d\times L}[/tex]

Put the value into the formula

[tex]d=\dfrac{27000}{7000\times3.14\times55}[/tex]

[tex]d=0.0223\ inch[/tex]

Hence, The diameter of a square link is 0.0233 inch.

Answer:

The answer is 3Q

Explanation:

The metal temperature increases in a linear way, we could get a difference between final and initial temperature

[tex]DT=FinalTemperature-InitialTemperature[/tex]

We get a temperature difference of 2 degrees per each heat addition.  If we add the same heat 3 times more, it will increase to 12 degrees

Answer:

The quantity of heat required to increase the temperature of the object from  6°C to 12°C is 3Q

Explanation:

Heat capacity is the quantity of heat required to increase the temperature of an object.

Q = mcΔθ

where;

Q is the quantity of heat

m is the mass of the object

c is specific heat capacity of the object

Δθ is change in temperature = T₂ - T₁

For the first sentence of this question;

Q = mc(6-4)

Q = mc(2)

Q = 2mc

For the second sentence of this question;

Let Q₂ be the quantity of heat required to increase the temperature of the object from  6°C to 12°C

Q₂ = mcΔθ

Q₂ = mc(12-6)

Q₂ = mc(6)

Q₂ = 6mc

Q₂ = 3(2mc)

Recall, Q = 2mc

Thus, Q₂ = 3Q

Answer:

The wave length is [tex]3.885\times10^{-13}\ m[/tex]

Explanation:

Given that,

Energy = 10 Mev

We need to calculate the wavelength

Using formula of debroglie wave length

[tex]\lambda=\dfrac{h}{\sqrt{2mE}}[/tex]

Where, h = Planck constant

E = energy

m = mass

Put the value into the formula

[tex]\lambda =\dfrac{6.634\times10^{-34}}{\sqrt{2\times9.11\times10^{-31}\times10\times10^{6}\times1.6\times10^{-19}}}[/tex]

[tex]\lambda=3.885\times10^{-13}\ m[/tex]

Hence, The wave length is [tex]3.885\times10^{-13}\ m[/tex]

Answer:

980 dyne

Explanation:

Volume = 1 cm^3, d = 0.92 g / cm^3, D = 1 g/cm^3

In the equilibrium condition, the buoyant force is equal to the weight of the block.

Buoyant force = Volume of block x density of water x g

Buoyant force = 1 x 1 x 980 = 980 dyne

Explanation:

(1) The force that exists between charged particles is electrostatic force. It is given by :

[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]

Where

q₁ and q₂ are charges

r is distance between charges

If the charge of a particle doubles, the electric force doubles. So, the correct option is (a) "It doubles".

(2) A charge exerts a negative force on another charge. Negative force denotes the force is attractive. It means that the charges are of opposite sign. So, the correct option is (c) "the charges are of opposite signs".

Answer:

The magnitude of the electric flux is [tex]3.53\ N-m^2/C[/tex]

Explanation:

Given that,

Electric field = 2.35 V/m

Angle = 25.0°

Area [tex]A= 1.65 m^2[/tex]

We need to calculate the flux

Using formula of the magnetic flux

[tex]\phi=E\cdot A[/tex]

[tex]\phi = EA\cos\theta[/tex]

Where,

A = area

E = electric field

Put the value into the formula

[tex]\phi=2.35\times1.65\times\cos 25^{\circ}[/tex]

[tex]\phi=2.35\times1.65\times0.91[/tex]

[tex]\phi=3.53\ N-m^2/C[/tex]

Hence, The magnitude of the electric flux is [tex]3.53\ N-m^2/C[/tex]

The speed of the car on Curve B is obtained by solving for the radius in Curve A's equation, then substituting that into the formula for Curve B. The formula involved is based on principles of circular motion and forces.

The subject of this question is Physics, specifically the principles of circular motion and the forces at play within. Given the information from Curve A, we can ascertain that the speed of the car on Curve B can be found using principles of physics. The formula to find the speed at which the car can travel around a banked curve without relying on friction is: v = sqrt[rgtan(Θ)]. Where v is the speed, r is the radius of the curve, g is the acceleration due to gravity, and Θ is the angle of the banked curve.

Since the radius is the same for both curves, and the speed is known for Curve A, we can set it up so: 19.1 = sqrt[r * 9.8 * tan(12.7)]. We then solve for r (the radius), and apply it to Curve B: v = sqrt[r * 9.8 * tan(15.1)]. By substituting in the value of r obtained from the first equation, we can calculate the speed for Curve B.

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To find the speed at which the car can travel around curve B without relying on friction, we can use the same formula but with a banked angle of 15.1°.

The speed at which a car can travel around a banked curve without relying on friction can be calculated using the ideal banking angle formula. The formula is given by v = √(g * r * tan(θ)), where v is the speed, g is the acceleration due to gravity, r is the radius of the curve, and θ is the banked angle. In this case, curve A is banked at 12.7° and the car can travel at a speed of 19.1 m/s. To find the speed at which the car can travel around curve B without relying on friction, we can use the same formula but with a banked angle of 15.1°.

Plugging in the values into the formula, v = √(9.8 * r * tan(15.1)). Since the radius is the same for both curves, we can solve for v: 19.1 = √(9.8 * r * tan(12.7))

Solving the equation for v, we find that the car can travel around curve B without relying on friction at a speed of approximately 21.2 m/s.

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(a) The velocity of the car-truck system after collision is 18.08 m/s.

(b) The increase in internal energy of the car and truck is 2,835,031.2 J.

The velocity of the system after collision is determined by applying the principle of conservation of linear momentum as shown below;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

2800(36) + 4000(-15) = vx(2800 + 4000)

40,800 = 6800vx

vx = 6 m/s

m₁u₁ + m₂u₂ = v(m₁ + m₂)

2800(0) + 4000(29) = vz(2800 + 4000)

116,000 = 6800vz

vz = 17.06 m/s

K.E(car) = ¹/₂mv²

K.E(car) = ¹/₂ x (2800) x (36)²

K.E(car) = 1,814,400 J

v(truck) = √(15² + 29²) = 32.65

K.E(truck) = ¹/₂ x (4000) x (32.65)²

K.E(truck) = 2,132,045 J

K.E(total) =  1,814,400 J + 2,132,045 J = 3,946,445 J

K.E =  ¹/₂(m₁ + m₂)v²

K.E = ¹/₂ x (2800 + 4000) x (18.08)²

K.E = 1,111,413.8 J

U = ΔK.E

U = 3,946,445 J - 1,111,413.8 J

U = 2,835,031.2 J

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Final answer:

To determine the velocity of the stuck-together car and truck after the collision, apply the principle of conservation of momentum. Consider the change in kinetic energy and the production of thermal energy and deformation.

Explanation:

To determine the velocity of the stuck-together car and truck just after the collision, we can apply the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision. We can find the final velocity by summing the momenta of the car and truck and dividing by their combined mass.

In this case, the car's momentum is the product of its mass and velocity, and the truck's momentum is the product of its mass and velocity. Adding these momenta together and dividing by the combined mass gives us the final velocity of the stuck-together car and truck just after the collision.

For the increase in internal energy of the car and truck, we need to consider the change in kinetic energy and the production of thermal energy and deformation. The change in kinetic energy can be calculated by finding the difference between the initial and final kinetic energies of the car and truck. The thermal energy and deformation depend on factors such as the materials involved and the severity of the collision.

Answer:

a) Maximum height reached above ground = 2.8 m

b) When he reaches maximum height he is 2 m far from end of the ramp.

Explanation:

a) We have equation of motion v²=u²+2as

   Considering vertical motion of skateboarder.  

   When he reaches maximum height,

          u = 6.6sin58 = 5.6 m/s

          a = -9.81 m/s²

          v = 0 m/s

  Substituting

         0²=5.6² + 2 x -9.81 x s

          s = 1.60 m

  Height above ground = 1.2 + 1.6 = 2.8 m

b) We have equation of motion v= u+at

   Considering vertical motion of skateboarder.  

   When he reaches maximum height,

          u = 6.6sin58 = 5.6 m/s

          a = -9.81 m/s²

          v = 0 m/s

  Substituting

         0= 5.6 - 9.81 x t

          t = 0.57s

  Now considering horizontal motion of skateboarder.  

  We have equation of motion s =ut + 0.5 at²

          u = 6.6cos58 = 3.50 m/s

          a = 0 m/s²

          t = 0.57  

  Substituting

         s =3.5 x 0.57 + 0.5 x 0 x 0.57²

         s = 2 m

  When he reaches maximum height he is 2 m far from end of the ramp.

The highest point reached by the skateboarder is 1.612 meters above the ground. When the skateboarder reaches the highest point, the horizontal distance from this point to the end of the ramp is  3.568 meters.

Given:

Initial velocity (v₀) = 6.6 m/s

Launch angle (θ) = 58°

Height of the ramp (h) = 1.2 m

Acceleration due to gravity (g) = 9.8 m/s²

(a) To find the maximum height reached by the skateboarder:

Δy = v₀y² / (2g)

v₀y = v₀ × sin(θ)

v₀y = 6.6 × sin(58°)

v₀y = 5.643 m/s

Δy = (5.643 )² / (2 × 9.8)

Δy ≈ 1.612 m

Therefore, the highest point reached by the skateboarder is 1.612 meters above the ground.

(b) The time of flight can be calculated using the equation:

t = 2 × v₀y / g

t = 2 × 5.643  / 9.8

t = 1.153 s

Δx = v₀x × t

First, we need to find the initial horizontal velocity (v₀x):

v₀x = v₀ × cos(θ)

v₀x = 6.6 × cos(58°)

v₀x = 3.099 m/s

Δx = 3.099 * 1.153

Δx = 3.568 m

Therefore, when the skateboarder reaches the highest point, the horizontal distance from this point to the end of the ramp is  3.568 meters.

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Answer:

The shear deformation is [tex]\Delta x=3.34\times10^{-6}\ m[/tex].

Explanation:

Given that,

Shearing force F = 600 N

Shear modulus [tex]S = 1\times10^{9}\ N/m^2[/tex]

length = 0.700 cm

diameter = 4.00 cm

We need to find the shear deformation

Using formula of shear modulus

[tex]S=\dfrac{Fl_{0}}{A\Delta x}[/tex]

[tex]\Delta x=\dfrac{Fl_{0}}{(\dfrac{\pi d^2}{4})S}[/tex]

[tex]\Delta x=\dfrac{4Fl_{0}}{\pi d^2 S}[/tex]

Put the value into the formula

[tex]\Delta x=\dfrac{4\times600\times0.700\times10^{-2}}{3.14\times1\times10^{9}\times(4.00\times10^{-2})^2}[/tex]

[tex]\Delta x=3.34\times10^{-6}\ m[/tex]

Hence, The shear deformation is [tex]\Delta x=3.34\times10^{-6}\ m[/tex].

The shear deformation experienced by the disc is calculated using a formula that takes into account the shear modulus, the force applied, and the cross-sectional area of the disk. The correct answer is found to be approximately 0.478 μm.

Solving this problem involves understanding the formula for shear deformation, which is the ratio of the applied force to the area of the disc over which it is applied, multiplied by the height of the disc and divided by the shear modulus.

First, we need to calculate the cross-sectional area of the disk. The formula for the area of a circle is πr², where r is the radius of the disc. Given the diameter of 4 cm, the radius is 2 cm or 0.02 m. So, the area = π * (0.02)² = 0.001256 m².

Substituting into the formula for shear deformation, we get τ = F / (G * A) which equals 600 N / (1x10^9 N/m² * 0.001256 m²) = 4.78x10^-7 m or approximately 0.478 μm.

This indicates that none of the initial answers are correct. The closest incorrect answer is 3 μm but the correct answer is 0.478 μm.

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By using equations from physics pertaining to projectile motion and manipulation of initial velocity, final velocity components, and combined final velocity, we can calculate the initial speed at which the rock was thrown and its velocity just before striking the ground.

The subject of this question is projectile motion, a branch of physics. Given that the horizontal range of the throw is equal to the height of the building, we can apply the equation for range in projectile motion: R = (v² sin 2α) / g, where R is the range (20m), v is velocity, α is the angle (53 degrees), and g is acceleration due to gravity (approx. 9.8 m/s²).

Firstly, solve the equation for v which gives v = sqrt(R * g / sin 2α). This gives the starting speed of the rock.

To find the final velocity just before hitting the ground, we need to find vertical and horizontal components of velocity. Vertical component can be obtained by using: v_f = sqrt(v_i² + 2*g*h), h is the height(20m). Horizontal component remains constant which is v_i*cosα. The final velocity is then, sqrt(v_h² + v_f²).

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Answer:

0.198 m/s

Explanation:

m1 = 0.7 kg, u1 = 0.15 m/s,

v1 = 0.08 m/s

m2 = 0.5 kg, u2 = 0.1 m/s

Let the speed of 0.5 kg is v2 after the collision.

By using the conservation of momentum

Momentum before collision = momentum after collision

m1 u1 + m2 u2 = m1 v1 + m2 v2

0.7 x 0.15 + 0.5 x 0.1

= 0.7 × 0.08 + 0.5 × v2

0.099 = 0.5 v2

v2 = 0.198 m/s

Answer:

Temperature, T = 1542.10 K

Explanation:

It is given that,

The black body radiation emitted from a furnace peaks at a wavelength of, [tex]\lambda=1.9\times 10^{-6}\ m[/tex]

We need to find the temperature inside the furnace. The relationship between the temperature and the wavelength is given by Wein's law i.e.

[tex]\lambda\propto \dfrac{1}{T}[/tex]

or

[tex]\lambda=\dfrac{b}{T}[/tex]

b = Wein's displacement constant

[tex]\lambda=\dfrac{2.93\times 10^{-3}}{T}[/tex]

[tex]T=\dfrac{2.93\times 10^{-3}}{\lambda}[/tex]

[tex]T=\dfrac{2.93\times 10^{-3}}{1.9\times 10^{-6}\ m}[/tex]

T = 1542.10 K

So, the temperature inside the furnace is 1542.10 K. Hence, this is the required solution.

Using Wien's Displacement Law, the temperature inside a furnace emitting radiation that peaks at a wavelength of 1.9 x 10^-6 m is approximately 1525 Kelvin.

The temperature of the furnace can be determined using Wien's Displacement Law, which states that the product of the temperature of a black body and the wavelength at which the radiation it emits is most intense is approximately equal to 2.898 x 10^-3 m.K. In this specific case, the peak wavelength of the energy emitted from the furnace is given as 1.9 x 10^-6 m. Therefore, the temperature inside the furnace can be calculated using the formula: T = Amax / λ, where Amax = 2.898 x 10^-3 m.K and λ = 1.9 x 10^-6 m. Calculating from the given formula, T ≈ 1525 K. Therefore, the temperature inside the furnace is approximately 1525 Kelvin.

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Answer:

[tex]\gamma = 2.97 \times 10^{-5} per ^0 C[/tex]

Explanation:

As we know by the theory of expansion the change in the volume of the object is directly proportional to change in temperature and initial volume.

So here we can say

[tex]\Delta V = V_0\gamma \Delta T[/tex]

here

[tex]\gamma[/tex] = coefficient of volume expansion

so we have

[tex]\gamma = \frac{\Delta V}{V_0 \Delta T}[/tex]

now plug in all values

[tex]\gamma = \frac{2.25 cm^3}{(541 cm^3)(273 - 133)}[/tex]

[tex]\gamma = 2.97 \times 10^{-5} per ^0 C[/tex]

Answer:

B) 20N.s is the correct answer

Explanation:

The formula for the impulse is given as:

Impulse = change in momentum

Impulse = mass × change in speed

Impulse = m × ΔV

Given:

initial speed  = 40m/s

Final speed = -60 m/s (Since the the ball will now move in the opposite direction after hitting the bat, the speed is negative)

mass = 0.20 kg

Thus, we have

Impulse = 0.20 × (40m/s - (-60)m/s)

Impulse = 0.20 × 100 = 20 kg-m/s or 20 N.s

The impulse experienced by the ball is,

[tex]\rm Impulse = 20\; Nsec[/tex]

Given :

Mass = 0.20 Kg

Initial Speed = 40 m/sec

Final Speed = -60 m/sec (minus sign shows that ball move in opposite direction)

Solution :

We know that the formula of Impulse is,

[tex]\rm Impulse = m\times \Delta V[/tex]

where, m is the mass and [tex]\rm \Delta V[/tex] is change in velocity.

[tex]\rm Impulse = 0.20\times(40-(-60))[/tex]

[tex]\rm Impulse = 20\; Nsec[/tex]

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