Heidi (39 kg) is walking her tiny chihuahua, Chaxi (5.60 kg), on the sidewalk. To encourage Chaxi along, Heidi pulls forward with a force of 9.55 N. Identify the correct reaction force in response to Heidi’s action force.

a. The friction is less than 660 N since the beam can be moved at a constant velocity.
b. There is no friction acting on the beam since it is accelerating.
c. The friction is equal to 660 N since the beam is not accelerating.
d. The friction is greater than 660 N since the beam is not in equilibrium.

Answer:

relative dating and radioactive dating are two methods in archeaology to determine the age of fossils and rocks

Explanation:

the act of or study of fossils is important for the determination of the kind of organism it represents how the organism lived and how it was preserved on the Earth’s surface over the past 4600000000 years

Answer:

480.2 m

Explanation:

The following data were obtained from the question:

Speed of sound (v) = 343 m/s.

Time (t) = 2.8 s

Distance (x) of the cliff =?

The distance of the cliff from the woman can be obtained as follow:

v = 2x /t

343 = 2x /2.8

Cross multiply

2x = 343 × 2.8

2x = 960.4

Divide both side by the coefficient of x i.e 2

x = 960.4/2

x = 480.2 m

Therefore, the cliff is 480.2 m away from the woman.

The distance should be 480.2 m

Since A woman standing before a cliff claps her hands, and 2.8s later she hears the echo. And, there is the velocity of 343 m/s

[tex]v = 2x \div t\\\\343 = 2x \div 2.8\\\\2x = 343 \times 2.8[/tex]

2x = 960.4

x = 480.2 m

Learn more: https://brainly.com/question/1504221?referrer=searchResults

The 400-foot altitude limit was put in place for the sake of airspace safety, and there is a risk to country security as well as the privacy of citizens.

Unmanned aerial vehicles (UAVs), sometimes known as drones, are used for a variety of jobs, from routine to extremely dangerous. These robotic-looking planes can be seen practically everywhere, from delivering groceries to your home to rescuing avalanche victims.

The 400-foot altitude restriction was ultimately implemented for airspace safety. Given the breadth of the airspace above 400 feet, the likelihood of a drone colliding with a human aircraft is exceedingly remote, but the consequences might be disastrous.

Any aerial vehicle that uses software to fly autonomously or that may be controlled remotely by a pilot is referred to as a drone. Numerous drones come equipped with cameras to gather visual data and propellers to stabilize flying paths. Drone technology has been incorporated into industries like videography, search and rescue, agriculture, and transportation.

When in uncontrolled (Class G) airspace, your drone must be flown 400 feet above the ground or less.

To know more about drones:

https://brainly.com/question/27753670

#SPJ2

Answer:

[tex]F = \frac{6.67408m_1 m_2}{10^{11}r^2}[/tex]

Explanation:

Given

[tex]Object_1 = m_1[/tex]

[tex]Object_2 = m_2[/tex]

[tex]Distance = r[/tex]

Required

Determine the force of attraction

This is calculated as:

[tex]F = \frac{GMm}{d^2}[/tex]

Where

M = mass of object 1

m = mass of object 2

d = distance

Where G = gravitational constant

[tex]G = 6.67408 * 10^{-11}\ m^3 kg^{-1} s^{-2}[/tex]

Substitute these values in

[tex]F = \frac{GMm}{d^2}[/tex]

[tex]F = \frac{6.67408 * 10^{-11} * m_1 * m_2}{r^2}[/tex]

[tex]F = \frac{6.67408 * m_1 * m_2* 10^{-11}}{r^2}[/tex]

[tex]F = \frac{6.67408m_1 m_2* 10^{-11}}{r^2}[/tex]

[tex]F = \frac{6.67408m_1 m_2}{10^{11}*r^2}[/tex]

[tex]F = \frac{6.67408m_1 m_2}{10^{11}r^2}[/tex]

Let w, n, p, and f denote the magnitudes of the 4 forces acting on the box.

• w = weight = 319 N

• n = normal force

• p = pushing force = 485 N

• f = friction = µ n, where µ is the coefficient of static friction

The net force on the box points in the direction that the box moves, which is to the right. In particular, this means the box is vertically in equilibrium. Split up the vectors into their vertical and horizontal components, and apply Newton's second law. (I take up and right to be the positive vertical and horizontal directions, respectively.)

• vertical:

p sin(-35°) + n - w = 0

and solving for n,

- (485 N) sin(35°) + n - 319 N = 0

n ≈ 597 N

• horizontal:

p cos(-35°) - f = m a

where a is the magnitude of the net acceleration on the box. Solve for a. Since f = µ n and m = w / g (where g = 9.80 m/s² is the mag. of the acc. due to gravity) we get

p cos(35°) - µ n = (w / g) a

(485 N) cos(35°) - µ (597 N) = (319 N) / (9.80 m/s²) a

a ≈ (12.2 - 18.3 µ) m/s²

(a) If µ = 0.57, then the net acceleration on the box is

a ≈ (12.2 - 18.3 • 0.57) m/s² ≈ 1.75 m/s²

so that the time t required to move the box 4 m is

4 m = 1/2 a t ²

t ≈ √((8 m) / (1.75 m/s²))

t ≈ 2.14 s

(b) The box does not move.

If µ = 0.75, then

a = (12.2 - 18.3 • 0.75) m/s² ≈ -1.55 m/s²

but a negative acc. here means the applied acc. points *opposite* the direction of movement, thus making the box move backward which doesn't make sense. The coefficient of friction is too large for the given applied force to get the box moving. With µ = 0.75, the frictional force to overcome has mag. f ≈ 448 N. But the given push contributes a horizontal force of (485 N) cos(-35°) ≈ 397 N. This mag. needs to be increased in order to get the box moving.

(a) The time taken to move the box 4 meters is 2.14 s.

(b) The box will decelerate when the coefficient of friction is 0.75 and cannot be moved to 4 meters forward.

The given parameters;

The mass of the books is calculated as;

[tex]m = \frac{W}{g} \\\\m = \frac{319}{9.8} \\\\m = 32.55 \ kg[/tex]

The normal force on the box is calculated as follows;

[tex]F_n = -W - Fsin\theta\\\\F_n = -319 - (485\times sin35)\\\\F_n = -597.18 \ N[/tex]

The frictional force when the coefficient of friction is 0.57;

[tex]F_f = \mu F_n\\\\F_f = 0.57 \times -597.18\\\\F_f = -340.39 \ N[/tex]

The acceleration of the box is calculated as follows;

[tex]F cos \theta - F_f = ma\\\\(485)cos(35) \ - 340.39 = 32.55 a\\\\56.899 = 32.55a\\\\a = \frac{56.899}{32.55} \\\\a = 1.75 \ m/s^2[/tex]

The time taken to move the box 4 meters is calculated as;

[tex]s = v_0t + \frac{1}{2} at^2\\\\s = 0 + \frac{1}{2} at^2\\\\t = \sqrt{\frac{2s}{a} } \\\\t = \sqrt{\frac{2\times 4}{1.75} } \\\\t = 2.14 \ s[/tex]

(b) The frictional force when the coefficient of friction is 0.75;

[tex]F_f = \mu F_n\\\\F_f = 0.75 \times -597.18\\\\F_f = -447.885 \ N[/tex]

The acceleration of the box is calculated as follows;

[tex]F cos \theta - F_f = ma\\\\(485)cos(35) \ -447.885 = 32.55 a\\\\-50.596 = 32.55a\\\\a = \frac{-50.596}{32.55} \\\\a = -1.55\ m/s^2[/tex]

Thus, the box will decelerate when the coefficient of friction is 0.75 and cannot be moved to 4 meters forward.

Learn more here:https://brainly.com/question/21684583

Type of meter would be connected on both sides of a resistor in a circuit is a a voltmeter , which measures potential difference .

A voltmeter is an instrument used for measuring the potential difference , or voltage between two points in an electrical circuit .

A voltmeter is always attached in a series combination and an ammeter (which measures current in a circuit ) always attached in parallel combination with the circuit.

Since , in question it is given that a meter would be connected on both sides of a resistor in a circuit that means it must be a series combination

hence , correct answer is B) a voltmeter , which measures potential difference .

learn more about voltmeter

https://brainly.com/question/8505839?referrer=searchResults

#SPJ3

Answer: True

Explanation:

Answer:

The magnitude of the bugs displacement is 3.87 m

Explanation:

An illustrative diagram for the scenario is given in the attachment below.

In the diagram, the bug's displacement is given by x. The diagram shows a right angle triangle with x as the hypotenuse. We can determine x from the Pythagorean theorem which states that " the square of the hypotenuse equals sum of squares of the other two sides". That is

x² = 2.25² + 3.15²

x² = 5.0625 + 9.9225

x² = 14.985

x = √14.985

x = 3.87 m

Hence, the magnitude of the bugs displacement is 3.87 m.

Answer:

The final velocity of the car is 26.65 m/s.

Explanation:

Given;

acceleration of the racecar, a = 6.5 m/s²

initial velocity of the car, u = 0

time of motion, t = 4.1 s

The final velocity of the car is given by;

v = u + at

where;

v is the final velocity of the car

suvstitute the givens

v = 0 + (6.5)(4.1)

v = 26.65 m/s.

Therefore, the final velocity of the car is 26.65 m/s.

Answer:

T' = 865.15 °C

Explanation:

In order for the copper collar to just slip on the steel shaft the, assuming are heated simultaneously, we must find the final parameters of both and equate them. Because the final diameters of both must be same for the slipping to occur.

FOR COPPER COLLAR:

dc' = dc(1 + ∝c*ΔT)

where,

dc' = final diameter of copper ring

dc = initial diameter of copper ring = 5.98 cm

∝c = coefficient of linear expansion for copper = 16 x 10⁻⁶ °C⁻¹

ΔT = Change in Temperature

Therefore,

dc' = (5.98 cm)[1 + (16 x 10⁻⁶ °C⁻¹)ΔT]   ------------- equation (1)

FOR STEEL SHAFT:

ds' = ds(1 + ∝s*ΔT)

where,

ds' = final diameter of steel shaft

ds = initial diameter of steel shaft = 6 cm

∝s = coefficient of linear expansion for steel = 12 x 10⁻⁶ °C⁻¹

ΔT = Change in Temperature

Therefore,

dc' = (6 cm)[1 + (12 x 10⁻⁶ °C⁻¹)ΔT]   ------------- equation (2)

Comparing equation (1) and equation (2):

(5.98 cm)[1 + (16 x 10⁻⁶ °C⁻¹)ΔT] = (6 cm)[1 + (12 x 10⁻⁶ °C⁻¹)ΔT]

(5.98 cm/6 cm)[1 + (16 x 10⁻⁶ °C⁻¹)ΔT] = [1 + (12 x 10⁻⁶ °C⁻¹)ΔT]

0.9967 + (1.59 x 10⁻⁵ °C⁻¹)ΔT = 1 + (12 x 10⁻⁶ °C⁻¹)ΔT

1 - 0.9967 = [(15.9 -12) x 10⁻⁶ °C⁻¹]ΔT

0.0033/3.9 x 10⁻⁶ °C⁻¹ = ΔT

ΔT = 846.15 °C

but,

ΔT = T' - T = T' - 19°C = 846.15°C

T' = 846.15 °C + 19 °C

T' = 865.15 °C

Answer:

The final velocity of the car is 16.893 m/s

The final velocity of the truck is 4.393 m/s

Explanation:

Given;

mass of the car, m₁ = 715 kg

mass of the truck, m₂ = 1490 kg

initial velocity of the car, u₁ = 0

initial velocity of the truck, u₂ = 12.5 m/s

let the final velocity of the car, = v₁

let the final velocity of the truck, = v₂

Apply the principle of conservation of linear momentum for elastic collision;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(715 x 0) + (1490 x 12.5) = 715v₁ + 1490v₂

18625 = 715v₁ + 1490v₂ -----equation (1)

Apply one-directional velocity formula;

u₁ + v₁ = u₂ + v₂

0 + v₁ = 12.5 + v₂

v₁ = 12.5 + v₂

Substitute v₁ into equation (1)

18625 = 715(12.5 + v₂) + 1490v₂

18625 =8937.5 + 715v₂ + 1490v₂

18625 - 8937.5 = 715v₂ + 1490v₂

9687.5 = 2205v₂

v₂ = 9687.5 / 2205

v₂ = 4.393 m/s

solve for v₁

v₁ = 12.5 + v₂

v₁ =  12.5 + 4.393

v₁ = 16.893 m/s

Answer:

The height above the top of the window is 1.44 m

Explanation:

Given;

time of motion, t = 0.32 s

height traveled at the given time, h = 2.2m

determine the initial velocity of the stone;

h = ut + ¹/₂gt²

2.2 = u(0.32) + ¹/₂ x 9.8 x 0.32²

2.2 = 0.32u + 0.502

0.32u = 2.2 - 0.502

0.32u = 1.698

u = 1.698 / 0.32

u = 5.31 m/s

This initial velocity on top of the window becomes the final velocity from the height above the window.

v² = u² + 2gh

where;

u is the initial velocity of the stone from the height above the window;

5.31² = 0 + (2 x 9.8)h

19.6h = 28.196

h = 28.196/19.6

h = 1.44 m

Therefore, the height above the top of the window is 1.44 m

Answer:

the velocity of the acorn

Explanation:

just do in in real life and see

Answer:

it is probably the velocity of the acorn

200 m

We are given:

Initial velocity of the plane (u) = 0 m/s

Final velocity of the plane (v) = 40 m/s

Time interval (t) = 10 seconds

Displacement of the plane (s) = x m

Solving for x:

Acceleration of the plane

v = u + at                                                     [First equation of motion]

40 = 0 + a(10)                                              [replacing known variables]

a = 4 m/s²                                                    [dividing both sides by 10]

Displacement of the Plane:

s = ut + 1/2 (at²)                                            [Second equation of motion]

s = (0)(10) + 1/2(4)(10)²                                  [replacing known variables]

s = 200 m

Hence, the Plane covers a distance of 200 m in the given time interval

Answer:

Work = Mass * Gravity * Height and is measured in Joules. Imagine you find a 2 -Kg book on the floor and lift it 0.75 meters and put it on a table. Remember, that “force” is simply a push or a pull. If you lift 100 kg of mass 1-meter, you will have done 980 Joules of work.

Explanation:

Answer:

High specific heat -> takes more energy to raise/lower object's temperature

Low specific heat -> takes less energy to raise/lower object's temperature

Explanation:

The specific heat capacity is the amount of heat required to raise the temperature of something per unit of mass.

A high specific heat value for an object means it takes more energy to raise or lower that object's temperature. A low specific heat value for an object means it does not take very much energy to heat or cool that object.

Answer:

the answer is a time your welcome

Answer:

(1)

Explanation:

Answer:

Magnetic field lines can be drawn by moving a small compass from point to point around a magnet. At each point, draw a short line in the direction of the compass needle.When opposite poles of two magnets are brought together, the magnetic field lines join together and become denser between the poles.

Explanation:

Answer:

1.23

Explanation:

[tex]{\underline{\pink{\textsf{\textbf{ Answer : }}}}}[/tex]

➩ 1.23 feet

[tex]{\underline{\purple{\textsf{\textbf{Explanation : }}}}}[/tex]

Given :

To find :

Solution :

As it is told that it's divided into four equal pieces

Therefore,

We must divide it by 4 to get the length of each piece.

So,

[tex] \sf \to \: \frac{4.92}{4} \\ \sf \to \: 1.23 \: feet \: ans.[/tex]

Answer:

v = 12.8 m/s

Explanation:

       [tex]p_{ox} = p_{fx} (1)[/tex]

       [tex]p_{oy} = p_{fy} (2)[/tex]

       [tex]p_{oy} = m_{t} * v_{ot} = 3520 kg* v_{ot} (3)[/tex]

       [tex]p_{fy} = m_{f} * v_{fy} = 5000 kg* 9.8 m/s * sin 66.9 = 45080 kg*m/s (4)[/tex]

       [tex]v_{ot} =\frac{45080kg*m/s}{3520kg} = 12.8 m/s (5)[/tex]

Answer:

A)828.8Hz

B)869.2Hz

Explanation:

Here is a complete question;

What frequency is received by a person watching an oncoming ambulance moving at 115 km/h and emitting a steady 753 Hz sound from its siren? Speed of sound is 345m/s

b. What frequency does she receive after the ambulance has passed?

Vs= speed of the ambulance

, We convert to m/s for unit consistency

= 115 km/h= 115km× 1000m/1m × 1hr/3600s= 31.94m/s

Dopler effect is when observed frequency of wave changes with respect to the source or when observed moves relative to transmitting medium can be expressed as

f'=[ (v + vo)/(v- vs)]*f

=[ (v )/(v- vs)]*f

The sign vo and vs depends on vthe direction of the velocity

f= frequency of ambulance siren= 753Hz

v= speed of sound in air= 345m/s

Vo= speed of observer= 0

A) we are to determine the f' of ambulance as heard by person as ambulance approaching.

To find the frequency f' observed by the person we use the expresion below

Then substitute the values

f'=[ (v )/(v- vs)]*f

=[ (345)/(345-31.94)]×753

= 828.8Hz

B)What frequency does she receive after the ambulance has passed?

To find the frequency f' observed by the person we use the expresion below

Then substitute the values

f'=[ (v )/(v + vs)]*f

=[ (345)/(345 + 31.94)]×753

= 869.2Hz

=

Answer:

false

Explanation:

[tex]\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}[/tex]

Actually Welcome to the Concept of the Kinematics in real world.

So, as given here, we have to find the Mass of the bus from the given momentum, so we get as,

P = m * V

momentum = mass * velocity

here, P= 152625 kgm/s and v= 11.1 m/s

so substituting we get as,

m = 152625 ÷ 11.1 => 13,750 kg

hence,the mass of the bus is 13,750 kg.

Answer:

To find the weight of something, simply multiply its mass by the value of the local gravitational field, and you get a result in newtons (N). For example, if your mass is 50 kg (about 110 pounds), then your weight is (50) (9.8). The point that must be overwhelmingly emphasized is that weight is a force.

Explanation:

Answer:

p₂ / p₁ = 2 (v₁ / v₂)

Explanation:

The moment is a very useful concept, since it is one of the quantities that is conserved during shocks and explosions, for which it had to be redefined to be consistent with special relativity,

         p = m v / √[1+ (v/c)² ]

for the case of speeds much lower than the speed of light this expression is close to

         p = m v

In this exercise they indicate that the moment of the second particle is twice the moment of the first, when their velocities are small

        p₂ = 2 p₁

       p₂/p₁ = 2

in consecuense

       m v₂ = 2 m v₁

       v₂ = 2 v₁

consider particles of equal mass.

By the time their speeds increase they enter the relativistic regime

        p₂ = mv₂ /√(1 + v₂² /c²)

        p₁ = m v₁ /√(1 + v₁² / c²)

let's look for the relationship between these two moments

       p₂ / p₁ = mv₂ / mv₁   [√ (1+ v₁² / c²) /√ (1 + v₂² / c²)

from the initial statement

      p₂ / p₁ = 2 √(c² + v₁²) / (c² + v₂²)

we take c from the root

      p₂ / p₁ = 2 √ [(1+ v₁²) / (1 + v₂²)]

this is the exact result, to have an approximate shape suppose that the velocities are much greater than 1

      p₂ / p₁ = 2 √ [v₁² / v₂²] = 2 √ [(v₁ / v₂)²]

      p₂ / p₁ = 2 (v₁ / v₂)

we see the value of the moment depends on the speed of the particles

Answer:

0.752 m/s

Explanation:

m1 = 3.00kg

u1 = 5.05m/s

m2 = 2.76kg

u2 = -3.66m/s

According to the law of conservation of momentum,

m1u1 + m2u2 = (m1+m2)v

3(5.05) + 2.76(-3.66) = (5.05+2.76)v

15.15 - 9.2736 = 7.81v

5.8764 = 7.81v

v = 5.8764/7.81

v = 0.752m/s

Answer: The rate at which an object moves towards a target is Speed

Explanation:

Rate is something that tells us amount of solmething that changes in one unit of time.

Speed is defined as the measure of the rate of movement of a body expressed either as the distance travelled divided by the time taken.

Arc is defined as the apparent path described above and below the horizon taken up by a celestial body.

Force is defined as a push or pull upon an object resulting from the object's interaction with another object.

Trajectory is defined as the path followed by an object moving under the action of given force.