You are driving in such a way that the car is accelerating at a constant rate in the positive direction. When you pass the first sign, you are traveling at 4 m/s. When you pass the second sign 50 m down the road, you note that the seconds indicator of your clock reads 45 seconds. You also note that your velocity is now 9 m/s.

Required:
a. What is your acceleration?
b. What was the clock’s seconds indicator reading when you passed the first sign?

Answer:

2) The bending of a light wave as it passes at an angle from one medium to another .

Hope it is helpful to you ☺️

Answer:

periodic

Explanation:

Answer:

The pressure is 2505 Pa.  

Explanation:

Height, h = 1.5 m

density of blood, d = 1000 kg/cubic meter

Gravity, g = 1.67 m/s^2

let the pressure is P.  

The pressure due to the fluid is given by

P = h d g

P = 1.5 x 1000 x 1.67

P = 2505 Pa

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 highest point in the trajectory occurs at the midpoint of the path. This highest point increases as the angle increases. At a 75° launch angle, the maximum height is approximately 76 meters. However, a further increase in launch angle beyond this 75° angle will increase the peak height even more.

Answer:

8m/s

Explanation:

Vavg= 16-0/2=8m/s

Answer:

[tex]\sigma=0.014\ C/m^2[/tex]

Explanation:

Given that,

The radius of sphere, r = 5 cm = 0.05 m

Net charge carries, q = 7.5 µC = 7.5 × 10⁻⁶ C

We need to find the surface charge density on the sphere. Net charge per unit area is called the surface charge density. So,

[tex]\sigma=\dfrac{7.5\times 10^{-6}}{\dfrac{4}{3}\pi \times (0.05)^3}\\\\=0.014\ C/m^2[/tex]

So, the surface charge density on the sphere is [tex]0.014\ C/m^2[/tex].

Answer:

1) attached below

2) assumption that the earth is spherical

Explanation:

1) Four illustrations of a globe

attached below

2) Reason for distortions at areas that do not fall on lines of tangency or secancy

The reason for distortion on areas outside the lines of tangency or secancy is because of the assumption that the earth is spherical which is not true hence map projections on the areas that fall on the lines of tangency do not experience distortion and are true

Answer:

The answer is "151.25 J and -547.64 J".

Explanation:

[tex]u = 0\\\\v = 2.35\ \frac{m}{sec}\\\\d = 5.0 \ m\\\\[/tex]

Using formula:

[tex]v^2 = u^2 + 2 \times a \times d\\\\2.35^2 = 0^2 + 2 \times a \times 5\\\\a = \frac{2.35^2}{10} \\\\[/tex]

   [tex]= 0.55 \ \frac{m}{sec^2}\\\\[/tex]

[tex]F_{net} = m \times a\\\\F_{net} = 55 \times 0.55 = 30.25\ N\\\\[/tex]

Calculating the Work by net force

[tex]W = F_{net}\times d\\\\W = 30.25 \times 5 = 151.25 \ J\\\\[/tex]

The above work is converted into thermal energy.

Now,

[tex]F_{net} = F_p - F_f\\\\F_p = 140 \ N\\\\F_f = u_k\times m \times g = u_k \times 55 \times 9.81\\\\F_f = 539.55 \times u_k\\\\30.25 = 140 - u_k \times 55 \times 9.81\\\\u_k = \frac{(140 - 30.25)}{(55\times 9.81)}\\\\uk = 0.203 = \text{Coefficient of friction}\\\\W_f = -F_f \times d\\\\W_f = -0.203 \times 55 \times 9.81 \times 5\\\\Work\ done\ by\ friction = -547.64 \ J[/tex]

Let the point where A is launched act as the origin, so that the horizontal positions at time t of the respective projectiles are

• A : x = (19.5 m/s) cos(30°) t

• B : x = 62.2 m + (19.5 m/s) cos(60°) t

These positions are the same at the moment one projectile is directly above the other, which happens for time t such that

(19.5 m/s) cos(30°) t = 62.2 m + (19.5 m/s) cos(60°) t

Solve for t :

(19.5 m/s) (cos(30°) - cos(60°)) t = 62.2 m

t = (62.2 m) / ((19.5 m/s) (cos(30°) - cos(60°))

t ≈ 8.71 s

Answer:

Look at work

Explanation:

Elastic Collision: Ki=Kf

M1=4.65kg

M2: 0.060kg

v1=5m/s

v2=0m/s

4.65*5+0.060*0=4.65*v1'+0.060*v2'

23.25+0=4.65v1'+0.060v2'

Also since it is an elastic collision we can use

v1+v1'=v2+v2'

4.65+v1'=v2'

4.65+v1'=v2'

Substitute into the earlier equation

23.25=4.65v1'+0.060(4.65+v1')

Expand

23.25=4.65v1'+0.279+0.06v1'

Solve for v1'

22.971=4.71v1'

v1'=4.88m/s

v2'=4.65+4.88=9.53m/s

Answer:

C.49 is yr ans...

Answer:b the force are equal and opposite each other

Explanation:

Answer:

t= 100s

Explanation:

use v=v0+at

plug in givens and solve for t

30=20+0.1*t

t= 100s

Answer:

Explanation:

Answer:

A) v_{f1} = -3.2 m / s,  B) LEFT , C) v_{f2} = -0.12 m / s,  D) LEFT

Explanation:

This is a collision exercise that can be solved using momentum conservation, for this we define a system formed by gliders, so that the forces during the collision are internal and the moment is conserved.

Let's use the subscript 1 for the lightest glider m1 = 0.160 kg and vo1 = 0.820 m / s

subscript 2 for the heaviest glider me² = 0.820 kg and vo2 = -2.27 m / s

Initial instant. Before the crash

          p₀ = m₁ v₀₁ + m₂ v₀₂

Final moment. After the crash

          p_f = m₁ v_{f1} + m₂ v_{f2}

          p₀ = p_f

          m₁ v₀₁ + m₂ v₀₂ = m₁ v_{f1} + m₂ v_{f2}

 as the shock is elastic, energy is conserved

         K₀ = K_f

         ½ m₁ v₀₁² + ½ m₂ v₀₂² = ½ m₁ [tex]v_{f1}^2[/tex] + ½ m₂ [tex]v_{f2}^2[/tex]

         m₁ (v₀₁² - v_{f1}²) = m₂ (v_{f2}² -v₀₂²)

let's make the relationship

         (a + b) (a-b) = a² -b²

         m₁ (v₀₁ + v_{f1}) (v₀₁-v+{f1}) = m₂ (v_{f2} + v₀₂) (v_{f2} -v₀₂)

let's write our two equations

         m₁ (v₀₁ -v_{f1}) = m₂ (v_(f2) - v₀₂)                                  (1)

         m₁ (v₀₁ + v_{f1}) (v₀₁-v_{f1}) = m₂ (v_{f2} + v₀₂) (v_{f2} -v₀₂)

we solve

         v₀₁ + v_{f2} = v_{f2} + v₀₂

we substitute in equation 1 and obtain

         M = m₁ + m₂

         [tex]v_{f1} = \frac{m_1-m_2}{M} v_o_1 + 2 \frac{m_2}{M} v_f_2[/tex]

         [tex]v_f_2 = \frac{2m_1}{M} v_o_1 + \frac{m_2-m_1}{M} v_o_2[/tex]vf2 = 2m1 / mm vo1 + m2-m1 / mm vo2

we calculate the values

         m₁ + m₂ = 0.160 +0.3000 = 0.46 kg

         v_{f1} = [tex]\frac{ 0.160 -0.300} {0.460} \ 0.820 + \frac{2 \ 0300}{0.460} \ (-2.27)[/tex]

         v_{f1} = -0,250 - 2,961

          v_{f1} = - 3,211 m / s

         v_{f2} = [tex]\frac{2 \ 0.160}{0.460} \ 0.820 + \frac{0.300 - 0.160}{0.460 } \ (-2.27)[/tex]

         v_{f2} = 0.570 - 0.6909

         v_{f2} = -0.12 m / s

now we can answer the different questions

A) v_{f1} = -3.2 m / s

B) the negative sign indicates that it moves to the left

C) v_{f2} = -0.12 m / s

D) the negative sign indicates that it moves to the LEFT

the simple answer is from 61kmph to 120kmph

Explanation:

no explanation is needed

Answer:

Collisions may take place between the reactants.

Explanation:

The collision frequency must be greater than the frequency factor for the reaction. A collision between the reactants must occur.

Answer:

13.94 rpm

Explanation:

Given that,

The diameter of the pulley, d = 14 foot

Radius, r = 7 foot

The linear velocity of the pulley, v = 14 mph = 20.53 ft/s

We need to find the angular velocity in rpm.

We know that, the relation between the linear velocity and the angular velocity is as follows :

[tex]v=r\omega\\\\\omega=\dfrac{v}{r}\\\\\omega=\dfrac{20.53}{14}\\\\\omega=1.46\ rad/s[/tex]

or

[tex]\omega=13.94\ rpm[/tex]

So, the angular velocity of the pulley is 13.94 rpm.

Answer:

The work done is 13680.8 J.

Explanation:

The work done can be calculated as follows:

[tex] W = F*d [/tex]              

Where:            

F: is the force                                                        

d: is the displacement = 40 m                                    

The force acting on the boulder is given by:

[tex] F = mgsin(\theta) [/tex]

Where:

m: is the mass = 100 kg

g: is the acceleration due to gravity = 10 m/s²

θ: is the angle = 20°      

Then, the work is:

[tex] W = mgsin(\theta)d = 100 kg*10 m/s^{2}*sin(20)*40 m = 13680.8 J [/tex]

Therefore, the work done is 13680.8 J.  

I hope it helps you!  

Answer:

0

Explanation:

a = dv/dt

if v is constant than the slope of the v graph will be 0, so dv/dt is 0

a= 0

Answer: 2.4 millimeters = 0.0024 meters

Explanation: A millimeter is 1/1000 of a meter. By diving 2.4 by 1000, you get 0.0024.

Answer:

false.

Explanation:

We know that for a wave that moves with velocity V, with a wavelength λ, and a frequency f, we have the relation:

V = λ*f

So, if the velocity is constant and we increase the frequency to:

f' > f

we will have a new wavelength λ'

Such that:

V = f'*λ'

And V = f*λ

Then we have:

f'*λ' = f*λ

Solvinf for λ', we get:

λ' =(f/f')*λ

And because:

f' > f

then:

(f/f') < 1

Then:

λ' =(f/f')*λ < λ

So, if we increase the frequency, we need to decrease the wavelength.

So, for higher frequency waves, we must have proportionally shorter wavelengths.

Then we can conclude that the given statement:

"or waves moving through the atmosphere at a constant velocity, higher frequency waves must have proportionally longer wavelengths"

is false.

Answer:The answer is C

Explanation:

If both forces are measured in Newtons, then the net force is

F = (18, 7, -12) N + (16, -10, 16) N = (34, -3, 4) N

The toolbox undergoes a displacement (i.e. change in position) in the direction of the vector

d = (17, 14, -1) m - (15, 14, -8) m = (2, 0, -9) m

The total work done by the astronauts on the toolbox is then

F • d = (34, -3, 4) N • (2, 0, -9) m = (68 + 0 - 36) N•m = 32 J

The work done by the two astronauts is equal to 96 J.

work done?Work done is defined as the product of force applied and the distance moved by the force.

The forces applied = 18+16 N, 7+ -10 N, and -12 + 16N

Forces = 34 N, -3 N, and 4N

Distances = (17 - 15, 14 - 14, -1 - - 8) m

Distances = 2, 0, 7

Work done = 34 × 2 + -3 × 0 + 4 × 7

Work done = 96 J

Therefore, the work done by the two astronauts is equal to 96 J.

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Speed= distance/time

Or time = distance/speed

According to your question

Speed=15m/s

and. Distance=1.2km. ,we must change kilometer in meter because given speed is in m/s

D= 1.2km = 1.2×1000m =1200meter

Time = distance/ speed

1200/15 =80second

Or. 1min and 20 sec will be your answer.

Answer:

paleontologist

Explanation:

Paleontologists are scientists that investigate the fossils of extinct life forms. Thus, the correct option is B.

A fossil is defined as the preserved trace, imprint, or proof of a once-living entity from a past geological era. Exoskeletons, bones, shells, impressions of animals or microbes in stone, objects preserved in amber, hair, petrified wood, and genetic traces are a few examples. The collection of all the fossils is called as the fossil record.

An organism from a past geologic era that has been preserved in the Earth's crust is referred to as a fossil. Paleontologists are scientists that investigate the fossils of extinct life forms. The intricate system of fossil records is the main source of information about the evolution of life on Earth.

Therefore, the correct option is B.

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Phytochemicals are compounds that are produced by plants ("phyto" means "plant"). They are found in fruits, vegetables, grains, beans, and other plants. Some of these phytochemicals are believed to protect cells from damage that could lead to cancer.