I need help with this physics question.

Answer:

2.21 N

Explanation:

The force in this case is the total mass multiplied by the acceleration due to gravity. You are not asked for the solution to be in terms of the torque which is the usual way to solve these problems. That's why you are not given where the fulcrum is.

The fulcrum feels F1 + F2 + 34 * 980

F2 = 141.7 * 980 = 138866

F1 = 50.3 * 980  =  49294

Ruler = 34 * 980=  33320

Total Force = 221480 The units here are dynes

I just saw in the middle of the question that g = 9.80

So the answer becomes 221480 / 1000 = 221.48   because we needed kg

And that answer becomes 221.48/100 2.21 because the force of gravity should be 9.8 not 980

The total force exerted on the fulcrum is

Answer:

a) acceleration

b) displacement

Explanation:

The velocity-time graph is a graph of velocity versus time. The velocity (m/s) would be on the Y-axis while time (s) would be on the X-axis.

a) The slope of a graph is given by: change in Y-axis/change in X-axis = ΔY/ΔX

In a velocity-time graph, ΔY = change in velocity and ΔX = change in time.

Hence, the slope of a velocity-time graph becomes: change in velocity/change in time.

Also, acceleration = change in velocity/change in time.

Hence, the slope of a velocity-time graph = acceleration.

b) Assuming that the area under a velocity-time graph is a rectangle, the area is given as:

Area of a rectangle = length x breadth

                                  = velocity x time (m/s x s)

Also, displacement = velocity x time (m)

Hence, the area under a velocity-time graph of a body would give the displacement of the body.

Answer:

Radio waves are used to carry satellite signals. These waves travel at 300,000 km/s (the speed of light). This means that a signal sent to a satellite 38,000 km away takes 0.13 s to reach the satellite and another 0.13 s for the return signal to be received back on Earth.

Explanation:

hope it help

Answer:

[tex]a_s=4.8\times 10^{-2}~m^2[/tex]

Explanation:

Given:

cross sectional area of the bone, [tex]a=4.8 \times 10^{-4} ~m^2[/tex]

factor of up-scaling the dimensions, [tex]s=10[/tex]

Since we need to find the upscaled area having two degrees of the dimension therefore the scaling factor gets squared for the area being it in 2-dimensions.

The scaled up area is:

[tex]a_s=a\times s^2[/tex]

[tex]a_s=[4.8 \times 10^{-4}]\times 10^2[/tex]

[tex]a_s=4.8\times 10^{-2}~m^2[/tex]

The area is defined as the space covered by an object in 2 d dimension. For a rectangle, it is a product of length and breadth. The new cross-section area will be 4.8×10⁻² m².

The area is defined as the space covered by an object in 2 d dimension. For a rectangle, it is a product of length and breadth. Its unit is m².

Given data in the problem

a is the crossectional area of conical bone = 4.8×10⁻⁴m².

s is the factor of up-scaling the dimensions =10

For two degrees of dimension, the upscaled area will be square of the given area.

The scaled-up area will be

[tex]\rm a_s=a\times s^2\\\\ a_s= 4.8\times10^{-4}\times {10}^2\\\\\ \rm a_s=4.8\times10^{-2}\;m^2[/tex]

Hence the new cross-section area will be 4.8×10⁻² m².

To learn more about the area refer to the link;

https://brainly.com/question/1631786

Explanation:

Explanation: It's because when he stop down from a moving bus his feet come at rest while the upper portion of his body is still in motion and he falls in the forward direction.

Answer:

20 seconds

Explanation:

We are given 2 givens in the first statement

v0=0 and a=5

And we are trying to find time needed to cover 1km or 1000m.

So we use

x-x0=v0t+1/2at²

Plug in givens

1000=0+2.5t²

solve for t

t²=400

t=20s

Answer:

I = 0.25 A

Explanation:

Given that,

Three 15 ohms and two 25 ohms light bulbs and a 24 V battery are connected in a series circuit.

In series combination, the equivalent resistance is given by :

[tex]R=R_1+R_2+R_3+....[/tex]

So,

[tex]R=15+15+15+25+25\\\\=95\ \Omega[/tex]

The current each resistor remains the same in series combination. It can be calculated using Ohm's law i.e.

V = IR

[tex]I=\dfrac{V}{R}\\\\I=\dfrac{24}{95}\\\\I=0.25\ A[/tex]

So, the current of 0.25 A passes through each bulb.

a. 4000

This has 4-digits.

Answer:

in my opinion letter d.

Explanation:

Sana pi tama

[tex]\large \mid \underline {\bf {{{\color{navy}{Leaf \: \: \: Chloroplast \: ...}}}}} \mid[/tex]

the simple answer is from 61kmph to 120kmph

Explanation:

no explanation is needed

Explanation:

a) photovoltaic cell is a semiconductor device and it converts light energy to electrical energy

b) burning of coal converts chemical energy to heat energy

c) rubbing of both hands against each other converts mechanical to heat energy

Answer:

a. solar cells

b.coal,wood,petroleum

c.rubbing ours palms

Answer:

  r ’= 4 r

Explanation:

Electric potential energy is

          U = [tex]k \frac{q_1q_2}{r_{12}}[/tex]k q1q2 / r12

in this exercise

          q₁ = q₂ = q

          U = k q² / r

for when the charge change

           U ’= k q’² / r’

indicate that

      q ’= 2q

      U ’= U

we substitute

           U = k (2q) ² / r ’

           U = 4 k q² / r ’

we substitute

           [tex]k \ \frac{q^2}{r} = 4 k \ \frac{q^2}{r'}[/tex]k q² / r = 4 k q² / r ’

           r ’= 4 r

Answer:

The wood block reaches a height of 4.249 meters above its starting point.

Explanation:

The block represents a non-conservative system, since friction between wood block and the ramp is dissipating energy. The final height that block can reach is determined by Principle of Energy Conservation and Work-Energy Theorem. Let suppose that initial height has a value of zero and please notice that maximum height reached by the block is when its speed is zero.

[tex]\frac{1}{2}\cdot m \cdot v^{2} = m \cdot g\cdot h + \mu_{k}\cdot m\cdot g\cdot s \cdot \sin \theta[/tex]

[tex]\frac{1}{2}\cdot v^{2} = g\cdot h + \mu_{k}\cdot g\cdot \left(\frac{h}{\sin \theta} \right)\cdot \sin \theta[/tex]

[tex]\frac{1}{2}\cdot v^{2} = g\cdot h +\mu_{k}\cdot g\cdot h[/tex]

[tex]\frac{1}{2}\cdot v^{2} = (1 +\mu_{k})\cdot g\cdot h[/tex]

[tex]h = \frac{v^{2}}{2\cdot (1 + \mu_{k})\cdot g}[/tex] (1)

Where:

[tex]h[/tex] - Maximum height of the wood block, in meters.

[tex]v[/tex] - Initial speed of the block, in meters per second.

[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, no unit.

[tex]g[/tex] - Gravitational acceleration, in meters per square second.

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

[tex]s[/tex] - Distance travelled by the wood block along the wooden ramp, in meters.

[tex]\theta[/tex] - Inclination of the wooden ramp, in sexagesimal degrees.

If we know that [tex]v = 10\,\frac{m}{s}[/tex], [tex]\mu_{k} = 0.20[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the height reached by the block above its starting point is:

[tex]h = \frac{\left(10\,\frac{m}{s} \right)^{2}}{2\cdot (1+0.20)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]

[tex]h = 4.249\,m[/tex]

The wood block reaches a height of 4.249 meters above its starting point.

Answer:

Approximately [tex]0.97\; \rm N[/tex]. This force would point away from the center of the square (to the left at [tex]45^\circ[/tex] above the horizontal direction.)

Explanation:

Coulomb's constant: [tex]k \approx 8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2}[/tex].

By Coulomb's Law, the electrostatic force between two point charges [tex]q_1[/tex] and [tex]q_2[/tex] that are separated by [tex]r[/tex] (vacuum) would be:

[tex]\displaystyle F = \frac{k \cdot q_1 \cdot q_2}{r^2}[/tex].

Consider the charge on the top-left corner of this square.

Apply Coulomb's Law to find the electrostatic force between this charge and the charge on the lower-left corner.

Convert quantities to standard units:

[tex]q_1 = q_2 = 3 \times 10^{-6}\; \rm C[/tex].

[tex]r = 0.40\; \rm m[/tex].

[tex]\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^2} \\ &\approx \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times (3 \times 10^{-6}\; \rm C)^{2}}{(0.40\; \rm m)^{2}} \\ &\approx 0.506\; \rm N\end{aligned}[/tex].

As the two charges are of the same sign, the electrostatic force on each charge would point away from the other charge. Hence, for the charge on the top-left corner of the square, the electrostatic force from the charge below it would point upwards.

Similarly, the charge to the right of this charge would exert an electrostatic force with the same magnitude (approximately [tex]0.506\; \rm N[/tex]) that points leftwards.

For the charge to the lower-right of the top-left charge, [tex]r = \sqrt{2} \times 0.40\; \rm m[/tex]. Therefore:

[tex]\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^2} \\ &\approx \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times (3 \times 10^{-6}\; \rm C)^{2}}{(\sqrt{2} \times 0.40\; \rm m)^{2}} \\ &\approx 0.253 \; \rm N\end{aligned}[/tex].

This force would point to the top-left of the top-left charge, which is [tex]45^\circ[/tex] above the horizontal direction. Decompose this force into two components that are normal to one another:

Consider the net force on the top-left charge in two components:

Combine these two components to find the magnitude of the net force on this charge:

[tex]\begin{aligned}F &= \sqrt{{F_x}^{2} + {F_y}^{2}} \\ &\approx \sqrt{0.685^2 + 0.685^2 }\; \rm N \\ &\approx 0.97\; \rm N\end{aligned}[/tex].

This force would point to the top-left of this charge (also at [tex]45^\circ[/tex] above the horizontal direction, away from the center of the square) because its horizontal and vertical components have the same magnitude.

Answer:

Angular displacement before it stops = 18 rev

Explanation:

Given:

Speed of fan w(i) = 6 rev/s

Speed of fan (Slow) ∝ = 1 rev/s

Final speed of fan w(f) = 0 rev/s

Find:

Angular displacement before it stops

Computation:

w(f)² = w(i) + 2∝θ

0² = 6² + 2(1)θ

0 = 36 + 2θ

2θ = -36

Angular displacement before it stops = -36 / 2

θ = -18

Angular displacement before it stops = 18 rev

Answer:

A point on the edge of the wheel will travel 199.563 radians at the given time.

Explanation:

Given;

initial angular velocity of the wheel; [tex]\omega _i = 245 \ rev/\min = 245\ \frac{rev}{\min} \times \frac{2\pi}{1\ rev} \times \frac{1 \ \min}{60 \ s} = 25.66 \ rad/s[/tex]

final angular velocity of the wheel;

[tex]\omega _f = 380 \ rev/\min = 380 \ \frac{rev}{\min} \times \frac{2\pi}{1\ rev} \times \frac{1 \ \min}{60 \ s} = 39.80 \ rad/s[/tex]

radius of the wheel, d/2 = (30 cm ) / 2 = 15 cm = 0.15 m

time of motion, t = 6.1 s

The angular distance traveled by the edge of the wheel is calculated as;

[tex]\theta = (\frac{\omega_f + \omega_i}{2} )t\\\\\theta = (\frac{39.8 + 25.66}{2} )\times 6.1\\\\\theta = 199.653 \ radian[/tex]

Therefore, a point on the edge of the wheel will travel 199.563 radians at the given time.

Answer:

a)[tex]|\Delta E|=4.58\: J[/tex]  

b)[tex]F=61.90\: N[/tex]

Explanation:

a)

We can use conservation of energy between these heights.

[tex]\Delta E=mgh_{2}-mgh_{1}=mg(h_{2}-h_{1})[/tex]  

[tex]\Delta E=0.608*9.81(0.6026-1.37)[/tex]

Therefore, the lost energy is:

[tex]|\Delta E|=4.58\: J[/tex]  

b)

The force acting along the distance create a work, these work is equal to the potential energy.

[tex]W=\Delta E[/tex]

[tex]F*d=mgh[/tex]

Let's solve it for F.

[tex]F=\frac{mgh}{d}[/tex]

[tex]F=\frac{0.608*9.81*1.37}{0.132}[/tex]

Therefore, the force is:

[tex]F=61.90\: N[/tex]

I hope is helps you!

Answer:

a. The average speed on her way to Grandmother's house is 48.08 mph

b. The average speed in the return trip is 50 mph.

Explanation:

The average speed (S) can be calculated as follows:

[tex] S = \frac{D}{T} [/tex]

Where:

D: is the total distance

T: is the total time

a. To find the total distance in her way to Grandmother's house, we need to find the total time:

[tex]T_{i} = t_{1_{i}} + t_{2_{i}} = \frac{d_{1_{i}}}{v_{1_{i}}} + \frac{d_{2_{i}}}{v_{2_{i}}}[/tex]

Where v is for velocity

[tex] T = \frac{d_{1_{i}}}{v_{1_{i}}} + \frac{d_{2_{i}}}{v_{2_{i}}} = \frac{(100/2) mi}{40.0 mph} + \frac{(100/2) mi}{60.0 mph} = 1.25 h + 0.83 h = 2.08 [/tex]    

Hence, the average speed on her way to Grandmother's house is:

[tex]S_{i} = \frac{D}{T_{i}} = \frac{100 mi}{2.08 h} = 48.08 mph[/tex]

b. Now, to calculate the average speed of the return trip we need to calculate the total time:                        

[tex]D = v_{1_{f}}\frac{T_{f}}{2} + v_{2_{f}}\frac{T_{f}}{2} = \frac{T_{f}}{2}(v_{1_{f}} + v_{2_{f}})[/tex]

[tex]100 mi = \frac{T_{f}}{2}(40 mph + 60 mph)[/tex]

[tex] T_{f} = \frac{200 mi}{40 mph + 60 mph} = 2 h [/tex]

Therefore, the average speed of the return trip is:

[tex]S_{f} = \frac{D}{T_{f}} = \frac{100 mi}{2 h} = 50 mph[/tex]

I hope it helps you!                                                      

Answer:

Explanation:

The density of pure water is 1 gram per 1 milliliter or one cubic cm. By knowing the density of water we can use it in dilution equations or to calculate the specific gravity of other solutions.

It can also help us determine what other substances are made of using the water displacement experiment. This is done by observing how much water is displaced when an object is submerged in the water. As long as you know the density of the water, the mass of the object being submerged and the volume of increase you can calculate the density of the object.

This was done by the great Archimedes in discovering what composed the kings crown.

Answer:

Yes

Explanation:

speed of truck = 20 km/h

Initially the car at rest.

maximum velocity of car = 40 km/h

When the truck and the car collide, the momentum of the truck transferred to car.

So, the car can attain the speed of 40 km/h.

Answer:

Well a meter stick has increments of a centimeter, and since 1 cm=0.01m he should record it as 4.02m(b)

Explanation:

Answer:

9.89 m/s.

Explanation:

Given that,

The radius of the circular arc, r = 25 m

The acceleration of the vehicle is 0.40 times the free-fall acceleration i.e.,a = 0.4(9.8) = 3.92 m/s²

Let v is the maximum speed at which you should drive through this turn. It can be solved as follows :

[tex]a=\dfrac{v^2}{r}\\\\v=\sqrt{ar} \\\\v=\sqrt{3.92\times 25} \\\\=9.89 m/s[/tex]

So, the maximum speed of the car should be 9.89 m/s.

30.1 N

Explanation:

Given:

[tex]W_1 = 16\:\text{N}[/tex]

[tex]W_2 = 8\:\text{N}[/tex]

Let's write the components of the net forces at the intersections. Note that the system is equilibrium so all the net forces are zero.

Forces involving W1:

[tex]x:\:\:\:-T_1 + T_3\cos \alpha = 0\:\: \\ \text{or}\:\:T_2 = T_3\cos \alpha\:\:\:\:\:(1)[/tex]

[tex]y:\:\:\:T_3\sin \alpha - W_1 = 0\:\:\: \\ \text{or}\:\:\:T_3\sin \alpha = W_1\:\:\:\:\:\:(2)[/tex]

Forces involving W2:

[tex]x:\:\:\:T_1\sin 53 - T_3\sin \alpha = W_2\:\:\:\:\:\:\:(3)[/tex]

[tex]y:\:\:\:T_4 - T_1\cos 53 - T_3\cos \alpha = 0\:\:\:\;(4)[/tex]

Substitute (2) into (3) and we get

[tex]T_1\sin 53 - W_1 = W_2[/tex]

Solving for [tex]T_1[/tex],

[tex]T_1 = \dfrac{W_1 + W_2}{\sin 53} = 30.1\:\text{N}[/tex]

Answer:

The change in entropy is 1.6 W/K.

Explanation:

Thickness, d = 0.5 cm

Area, A = 1 m^2

T = 25°C

T' = - 10°C

Coefficient of thermal conductivity of glass, K = 0.8 W/mK

The change in entropy is given by

S = Q/T

Here,

[tex]S =\frac{Q}{T}\\\\S = \frac{K A (T - T')}{d(T - T')}\\\\S = \frac{0.8\times 1}{0.5} = 1.6 W/K[/tex]

Answer:

(a) t = 0 s

(b) t = 0 s, 30 s, 55 s

(c) t = 40 s to t = 60 s

(d) t = 10 s to t = 15 s

(e) a = 6 m/s^2

Explanation:

(a) The car is at starting position at t = 0 s and v = 0 m/s.

(b) The velocity of car is zero when the time is t = 0 s, 30 s and 55 s.

(c) from t = 40 s to 60 s the car is moving in the negative direction.

(d) The fastest speed is 60m/s from t = 10 s to t = 15 s.

(e) The slope of the velocity time graph gives acceleration.

a = (60 - 0) / (10 - 0) = 6 m/s^2

Answer:

a

Explanation:

Answer:

the answer is c

'

Explanation: