A car moving at 41.35 m/s hits a brick wall and stops in 0.140 s. What is the
acceleration of the car?

Answer:

Please find attached, the required wave drawn with MS Excel

Explanation:

Functions that represent waves is given as follows

A general form of the wave equation is A·sin(B·x) + D

Where;

B = 2·π/T

T = The period of the wave = 1/f

D = The vertical shift of the wave = 0

A = The amplitude of the wave = 1 for sine wave

v = The wave velocity

λ = The wavelength of the wave

f = The frequency of the wave

v = f·λ

At constant v, λ ∝ 1/f  

∴ λ ∝ T

Where T = 3, we have;

B = 2·π/T

∴ B = 2·π/3

Therefore, we have the wave with an amplitude of 1 cm, and wavelength, 3 cm, given as follows

y = sin((2·π/3)·x)

Plotting the above wave with MS Excel, we can get the attached wave

Initial velocity=10m/s=u

Final velocity=v=15m/s

Time=t=2s

[tex]\boxed{\sf Acceleration=\dfrac{v-u}{t}}[/tex]

[tex]\\ \sf\longmapsto Acceleration=\dfrac{15-10}{2}[/tex]

[tex]\\ \sf\longmapsto Acceleration=\dfrac{5}{2}[/tex]

[tex]\\ \sf\longmapsto Acceleration=2.5m/s^2[/tex]

Answer: a = 2.5 cm/s²

Explanation:

Acceleration = (Final velocity - Initial Velocity)/time taken

a = v-u/t

Initial velocity = 10 cm/s

Final velocity = 15 cm/s

Time = 2 seconds

a = (15-10)/2

a = 5/2

a = 2.5 cm/s²

Therefore the acceleration is 2.5 cm/s²

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In free fall

[tex]\boxed{\sf s=-\dfrac{1}{2}gt^2}[/tex]

[tex]\\ \sf\longmapsto s=-\dfrac{1}{2}\times 9.8(1)^2[/tex]

[tex]\\ \sf\longmapsto s=-4.9(1)[/tex]

[tex]\\ \sf\longmapsto s=-4.9m[/tex]

[tex]\\ \sf\longmapsto s=4.9m[/tex]

Answer:

Explanation:

High concentrations of hydrogen ions yield a low pH (acidic substances), whereas low levels of hydrogen ions result in a high pH (basic substances). The overall concentration of hydrogen ions is inversely related to its pH and can be measured on the pH scale

Answer:

The SI unit of upthrust is Newton(N).

The SI unit of preesure is Pascal(P).

Thank You

Maximum error in the result of the sum of measurement is equal to the sum absolute error of the individual observed measurements

(a) The maximum error in D is 0.09

(b) The standard error in D is approximately 0.034

The procedure for arriving at the above values is as follows;

The given measurements and the sited errors are;

A = 15.21 + 0.01

B = 10.82 + 0.05

C = 11.00 + 0.03

D = A + B + C

(a) Required parameter;

To calculate the maximum error in D

The equation for the propagation of error in addition is presented as follows;

Given that we have;

x = a + b

Therefore;

x + ±Δx = (a ± Δa) + (b ± Δb) = a + b ± (Δa + Δb)

∴ Δx = Δa + Δb

From the above formula, we have;

Where;

D = A + B + C

The maximum error in D = The sum of the maximum error in A, B, C

∴ The maximum error in D = 0.01 + 0.05 + 0.03 = 0.09

(b) Required parameter:

To find the standard error in D

The standard error is the sampling distribution's standard deviation, SD

Variance = SD²

The combined variance, SD² = The sum of the squares of individual standard deviations

Given that the standard errors represents the standard deviation, we get;

The combined variance, SD² = 0.01² + 0.05² + 0.03²

The combined variance, SD = √(0.01² + 0.05² + 0.03²) = 0.059

[tex]Standard \ error = \dfrac{SD}{\sqrt{n} }[/tex]

Where n = 3, for the three measurement, we get;

[tex]Standard \ error = \dfrac{\sqrt{0.01^2 + 0.05^2 + 0.03^2} }{\sqrt{3} } \approx 0.034[/tex]

The standard error in D is approximately 0.034

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The answer is definitely Potassium

12 x cos 50 = 7.713451316...

Answer:

a. Effort = 960 Newton

b. Mechanical advantage (M.A) = 0.625

c. Velocity ratio (V.R) = 1.67

Explanation:

Given the following data;

Conversion:

100 cm = 1 m

X cm = 0.75 m

Cross-multiplying, we have;

X = 0.75 * 100 = 75 cm

First of all, we would find the effort arm;

Effort arm = length of crow bar - length of load arm

Effort arm = 200 - 75

Effort arm = 125 cm

Next, we would determine the mechanical advantage (M.A) of the crow bar;

[tex] M.A = \frac {Effort \; arm}{Load \; arm} [/tex]

Substituting the values into the formula, we have;

[tex] M.A = \frac {125}{200} [/tex]

M.A = 0.625

To find the effort of the crow bar;

[tex] M.A = \frac {Load}{Effort} [/tex]

Making "effort" the subject of formula, we have;

[tex] Effort = \frac {Load}{M.A} [/tex]

[tex] Effort = \frac {600}{0.625} [/tex]

Effort = 960 Newton

Lastly, we would determine the velocity ratio (V.R);

[tex] V.R = \frac {length \; of \; effort \; arm}{length \; of \; load \; arm} [/tex]

[tex] V.R = \frac {125}{75} [/tex]

V.R = 1.67

Answer:

2ms-¹ means that the body under consideration moves 2m in a second, and may be it will continue to move 2m in every 1 second, if there's no external unbalanced force acting on that body (those forces do include frictional forces). mark its brainlist plz. Kaneppeleqw and 6 more users found this answer helpful. Thanks 3.

Answer:

that the student has travels 2 meters every 1 second that passes

Answer:

Density is how compact something is. The relationship is M/V=D (Mass divided by Volume equals Density).

Explanation:

WHAT IS DENSITY:

Density is the degree of compactness of a substance.

EXAMPLE:

"a reduction in bone density"

FORMULA OF DENSITY:

The formula for density is d = M/V, where d is density, M is mass, and V is volume.

Accident: Get burned

Prevention: turn off the burner.

Answer:

the product of mass and velocity

....in my syllabus

Answer:

4kg×5gm^3=60

Explanation:

the object if heavy

Answer:

A. DT is given by Q= MCs DT

m = mass of the substances

Cs= is it's specific heat capacity

Ck= Q

Mk ×DTk

=250 × 9 × 5

129

=Dt = 180.1085271

answer is 180degree C.

Explanation:

B. = 25×10 ×100

1.082

=2500

1.082

= 23105.360 g/kj.

The final temperature is 180 degree. and the specific heat of the metal is 23105.360 g/kj.

Q = m . C . ΔT

Q = heat; m = mass; C is the specific heat and

ΔT = Final T° - Initial T°

Q = C lat . m

Q = Heat

m = mass

C lar = Latent heat of fusion

A) DT is given by Q= M Cs DT

where, m = mass of the substances

Cs= is it's specific heat capacity

Ck= Q

Mk × DTk

=250 × 9 × 5

129 =Dt = 180.1085271

Thus, the final temperature is 180 degree.

B) We heat a 25–g sample of metal from 10 °C to 100 °C. 1.082 kJ of energy is added to it by heat = 25×10 ×100

=2500

1.082

Q = 23105.360 g/kj

Hence, the specific heat of the metal is 23105.360 g/kj.

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Answer:

a=3.4m because of the m

b=1day because 86400=a day

Answer:

It is correct to say that if the human race continues to use energy as it is now, without change, we will witness negative environmental impacts around the world in this century.

As a concrete example, we can cite the means of transport that use fossil fuels, such as cars and buses, which release polluting gases into the atmospheric layer and cause the greenhouse effect, contributing to global warming.

To solve these problems, it is necessary to raise the awareness of individuals, so that there is more and more interest and search for environmentally responsible solutions, such as the large-scale production of electric cars, which do not pollute the environment.

Yes it is the measurement in science

The "second" is the base unit of time in both systems.

Answer:

option c. 21.0

Explanation:

It was given that to find 3 significant figures. So the answer is 21.0

Explanation:

based on the above information

1.A

2.B

3. C

Answer:

Ke²

Explanation:

So,

q1 = e

q2 = e

r = 1m

By coulumb's law,

F = K (q1q2/r²)

F = K (e)(e)/(1)²  

F = Ke²  

Option(a)

Answer:

[tex]\boxed {\boxed {\sf 15 \ m/s \ or \ 15 \ m*s^{-1}}}[/tex]

Explanation:

We are asked to find the final velocity. We are given the acceleration, time, and initial velocity, so we can use the following kinematics formula.

[tex]v_f= v_i+ at[/tex]

In this formula, [tex]v_f[/tex] is the final velocity, [tex]v_i[/tex] is the initial velocity, [tex]a[/tex] is the acceleration, and [tex]t[/tex] is the time.

The bicycle has an initial velocity of 5.0 m *s⁻¹ or m/s, acceleration of 2 m/s², and a time of 5 seconds.

[tex]\bullet \ v_i = 5.0 \ m/s \\\bullet \ a= 2\ m/s^2\\\bullet \ t= 5 \ s[/tex]

Substitute the values into the formula.

[tex]v_f=5.0 \ m/s + ( 2\ m/s^2 * 5 \ s)[/tex]

Solve inside the parentheses.

[tex]v_f= 5.0 \ m/s + (10 \ m/s)[/tex]

Add.

[tex]v_f= 15 \ m/s[/tex]

The units can also be written as:

[tex]v_f= 15 \ m*s^{-1}[/tex]

The bicycle's final velocity is 15 meters per second.

Answer:

The speed of the comet at the surface of the star is approximately 1,208,694.7 m/s

Explanation:

Question parameter obtained online; The mass of the star, M = 5 × 10³¹ kg

Explanation;

The given mass of the comet, m = 2 × 10⁸ kg

The initial velocity of the comet, v → 0

The distance of the comet from the star, d = 700,000,000 km

The gravitational potential at d = G·M·m/d

The kinetic energy of the comet, K.E. = m·v²/2

The kinetic energy of the comet at d = m·(0)²/2 = 0

The gravitational potential at the surface of the star, R = G·M·m/R

The kinetic energy of the comet at the surface of the star, R = m·(v)²/2 = 0

Where;

M = The mass of the star = 5 × 10³¹ kg

[tex]M_{Sun}[/tex] = The mass of the Sun = 1.989 × 10³⁰ kg

M/[tex]M_{Sun}[/tex] = 5 × 10³¹/(1.989 × 10³⁰) ≈ 25

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

R = The radius of the star

Therefore, we have;

m·(0)²/2 - G·M·m/d = m·v²/2 - G·M·m/R

∴ v = √((G·M·m/R - G·M·m/d)×2/m) = √(2·G·M(1/R - 1/d))

Therefore; v = (2 × 6.67430 × 10⁻¹¹ × 5 × 10³¹ × (1/R - 1/700,000,000,000))

v = 81696389149.1×√(1/R - 1/700,000,000,000).

The speed of the comet at the surface of the star, v = 81696389149.1×√(1/R - 1/700,000,000,000)

The mass radius relationship is given as follows;

[tex]\dfrac{R}{R_{Sun}} = 1.30 \times \left(\dfrac{M}{M_{Sun}} \right)^{\dfrac{1}{2} }[/tex]

[tex]R = R_{Sun} \times 1.30 \times \left(\dfrac{M}{M_{Sun}} \right)^{\dfrac{1}{2} }[/tex]

The radius of the Sun = 696,340,000 M

∴ R ≈ 696,340,000 × 1.3 × √(25.14) = 4538865694.76

R = 4538865694.76 m

v = 81696389149.1×√(1/4538865694.76 - 1/700,000,000,000) ≈ 1208694.7  m/s

Answer:

condensation - thermal energy removed

freezing -thermal energy removed

deposition - thermal energy removed

sublimation - thermal energy added

evaporation - thermal energy added

melting - thermal energy added

Explanation:

Thermal energy is heat energy. Processes in which heat is added involve the addition of thermal energy while processes in which heat energy is removed involves removal of thermal energy.

Condensation involves a change from gas to liquid, freezing involves a change from liquid to solid while deposition involves the settling of mobile particles at a place. All these processes involve a decrease in energy of particles.

On the other hand, sublimation is a direct change from solid to gas, melting involves a change from solid to liquid while evaporation involves a change from liquid to gas. All these processes occur when energy is added to the particles in a system.

Answer:

condensation - thermal energy removed

freezing -thermal energy removed

deposition - thermal energy removed

sublimation - thermal energy added

evaporation - thermal energy added

melting - thermal energy added

Question 1; The elongation of the steel is approximately 0.3123 mm

Question 2; The percentage the density of water increased in the deepest

ocean is approximately 6.4%

The strategy of obtaining the above solution is presented as follows;

Que. 1; The given parameters are;

The mass of the suspended block, m = 10 kg

The length of the steel, l = 2 m

The radius of the steel, r = 1 mm = 1 × 10⁻³ m

The modulus of elasticity of steel, E = 200 GPa = 200 × 10⁹ Pa

The stress, σ, on the steel due to the mass, m, is given as follows;

[tex]\mathbf{\sigma = \dfrac{F}{A}}[/tex]

Where;

F = The force acting on the steel = The weight of the mass

A = The cross sectional area of the steel = π·r²

∴ F = 10 kg × 9.81 m/s² = 98.1 N

A = π × (1 × 10⁻³)² = 3.14159 × 10⁻⁶ m²

Therefore;

σ = 98.1 N/(3.14159 × 10⁻⁶ m²) ≈ 31,226,226.2 Pa

We have;

[tex]\mathbf{ E = \dfrac{\sigma}{\epsilon}}[/tex]

From which we have;

[tex]\epsilon = \dfrac{\sigma}{E}[/tex]

Where;

∈ = The tensile strain = Δl/l

Δl = The elongation of the steel

Therefore;

∈ = 31,226,226.2/(200 × 10^9) = 0.00015613113

∴ Δl = 0.00015613113 × 2 m = 0.00031226226 m = 0.31226226 mm

The elongation of the steel, Δl = 0.31226226 mm ≈ 0.3123 mm

Question 2

The given parameters are;

The change in pressure per unit depth, Δp = 1.0 atm per 10 meters

The depth of the ocean = 12 km = 12,000 m

The compressibility = 5.0 × 10⁻⁵

The formula for compressibility, C, is presented as follows;

[tex]C = \dfrac{1}{V} \times \dfrac{\partial V}{\partial P}[/tex]

The change in pressure, [tex]\partial P[/tex] = 12,000 m × 1.0 atm/(10 m) = 1,200 atm

For a unit volume, V = 1 m³

We get;

[tex]5 \times 10^{-5} = \dfrac{1}{1} \times \dfrac{\partial V}{1,200}[/tex]

[tex]\partial V[/tex] = 5 × 10⁻⁵ m³/(atm) × 1,200 = 0.06 m³

The volume occupied 1 m³ at 12,000 km depth = V - [tex]\partial V[/tex]

∴ The volume occupied 1 m³ at 12,000 km depth = 1 m³ - 0.06 m³ = 0.94 m³

The percentage density increase, [tex]\partial[/tex]ρ% = (m/0.94 - m/1)/m/1 × 100

∴ (1/0.94 - 1/1)/1/1 × 100 ≈ 6.4%

The percentage increase in density  ≈ 6.4%

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Ans

It is being pulled toward the object by the objects gravity

Gravity:-

Answer:

the scientific study of measurement.

This question can be solved using the equations of motion. There are two scenarios where the equations of motion can be used. The first scenario is the free-fall motion of the piece of bread. The second scenario is the uniformly accelerated motion of the bird.

The acceleration of the bird is  "a = 26.13 m/s²".

First, we will calculate the time taken by the bread to fall 3 m. Using the second equation of motion for this free-fall motion:

[tex]h = v_it + \frac{1}{2}gt^2[/tex]

where,

h = height fall = 3 m

vi = initial velocity = 0 m/s

g = acceleration due to gravity = 9.8 m/s²

t = time taken = ?

Therefore,

[tex]3\ m = (0\ m/s)t+\frac{1}{2}(9.8\ m/s^2)t^2\\t = \sqrt{\frac{(3\ m)(2)}{9.8\ m/s^2}}\\\\t = 0.78\ s[/tex]

The bird took the same time to catch the bread. Now applying the second equation of motion to the bird's motion:

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

where,

s = distance covered by the bird = 5 m + 3 m = 8 m

vi = initial velocity of the bird = 0 m/s

a = acceleration of the bird = ?

t = time taken = 0.78 s

Therefore, using these values we get:

[tex]8\ m = (0\ m/s)(0.78\ s)+\frac{1}{2}a(0.78\ s)^2\\\\a = \frac{16\ m}{(0.78\ s)^2}[/tex]

a = 26.13 m/s²

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