A greater applied force is required to move an object with a greater mass than one with a smaller mass.
True
False

Answer:

The brightness of a light depends on the amplitude of the light wave, which is the extent the waves moves from their equilibrium position. The brightness is also related to the amount of light that is emitted or reflected by an object

Therefore, compared to dim light, light that look bright have a higher amplitude and emit or reflect more light energy (photons)

Explanation:

Answer:

A.

Explanation:

The pad extends the time so the impulse changes.

Answer:

A. The pad extends the time so the impulse changes

Explanation:

in term of physics it has same mass:)

Answer:

2a.

a=1.13ms^-2

2b.

S=277m

2c.

V=27.7ms-¹

Explanation:

Initial Velocity (U)=22m/s¹

Final Velocity (V)=43m/s²

Time(t) =18.6s

a. a=V-U/t

a=43-22/18.6

a=1.129

a=1.13m/s²

2b.

S=ut+1/2 at²

s=22(10)+1/2×1.13(10)²

s= 220+0.57(10)²

s= 220+0.57(100)

s= 220+57

s=277m

2c.

V=U+AT

V=22+1.13(5)

V=22+5.65

V=22+5.7

V=27.7m/s¹

Answer:

(a) 37.5 kg

(b) 4

Explanation:

Force, F = 150 N

kinetic friction coefficient = 0.15

(a) acceleration, a = 2.53 m/s^2

According to the newton's second law

Net force = mass x acceleration

F - friction force = m a

150 - 0.15 x m g = m a

150 = m (2.53 + 0.15 x 9.8)

m = 37.5 kg

(b) As the block moves with the constant speed so the applied force becomes the friction force.

[tex]F = \mu m g \\\\150 = \mu\times 37.5\\\\\mu = 4[/tex]

Answer:

Explanation:

They traveled this distance in 2 parts, essentially. Part 1 had an average speed for a certain number of hours, part 2 had an average speed for a certain number of hours, and those 2 parts taken together took them a distance of 364 km. In equation form, that looks like this:

km/hr part 1 + km/hr part 2 = 364 km

Now we need to find each part on the left side of that equation. Part 1 first:

We traveled 65 km/hr for 2 hours, so that took us

[tex]65\frac{km}{hr}*2hr[/tex] and canceling out the hour label, we have that in part 1 we got

65(2) = 130 km. Good. Now onto the second part, where our unknown is.

We traveled 78 km/hr the second part for x hours, so that took us

[tex]78\frac{km}{hr}*xhr[/tex] and canceling out the hour label, we have that in part 2 we got

78x km. Now we can fill in the main equation (the one in bold print)

130 km + 78x km = 364 km and subtracting 130 km from both sides:

78x km = 234 km and dividing by 78 km:

x = 3 hours. Part 2 took 3 hours. Part 1 took 2 hours, so the whole trip took 5 hours.

Answer:

Gravitational potential

Explanation:

Any object that is not on the surface of the Earth, but at a height instead has potential energy. Eventually, this can become kinetic energy once the object falls.

When an object is in a gravitational field, it has energy in its gravitational potential energy store.

Answer:

a is your answer.............

Answer:

2000g

Explanation:

volume=mass/density

=2000/2.5

=800cm³

mass=density×volume

=800×2.5

=2000g

I hope this helps you ^-^

Answer:

f=ma.......m=f/a......m=20kg

Answer: If we leave liquid butane in a bowl on a table in a room where the temperature is [tex]24^{o}C[/tex], butane will evaporate.

Explanation:

A temperature at which the the liquid and gaseous phase of a substance of a substance are present in equilibrium with each other is called boiling point.

For example, the boiling point of butane is -1.5 degree Celsius.

This means that at a temperature above -1.5 degree Celsius, butane will exist is gaseous state. That is, at a temperature of 24 degree Celsius butane will evaporate.

Thus, we can conclude that if we leave liquid butane in a bowl on a table in a room where the temperature is [tex]24^{o}C[/tex], butane will evaporate.

Answer:

Densities increase down the group

MP and BP decrease down the group

Softness increased going down the group

Speed of reacting increases going down the group

Answer:

Explanation:

Use the kinematic equation.

This equation can be derived from [tex]f=ma[/tex], but we can just memorize, or look them up when needed as it saves us time.

Now we can plug our measurements into each variable to solve for acceleration.

Subtract 8m/s from both sides.

Divide by 5 seconds. Left with acceleration in terms of [tex]\frac{m}{s^2}[/tex]

Answer:

La longitud de la varilla de cobre es de 1.201 metros a una temperatura de 100 °C.

Explanation:

Asumiendo que la varilla de cobre experimenta deformaciones muy pequeñas y que las deformaciones no longitudinales son despreciables con respecto a las deformaciones longitudinales, la deformación longitudinal de la varilla se estima mediante la siguiente fórmula:

[tex]l_{f} = l_{o}\cdot [1+\alpha \cdot (T_{f}-T_{o})][/tex] (1)

Donde:

[tex]l_{o}[/tex] - Longitud inicial de la varilla, en metros.

[tex]\alpha[/tex] - Coeficiente de dilatación, en [tex]^{\circ}C^{-1}[/tex].

[tex]T_{o}[/tex] - Temperatura inicial de la varilla, en grados Celsius.

[tex]T_{f}[/tex] - Temperatura final de la varilla, en grados Celsius.

Si sabemos que [tex]l_{o} = 1.20\,m[/tex], [tex]\alpha = 1.4\times 10^{-5}\,^{\circ}C^{-1}[/tex], [tex]T_{o} = 18\,^{\circ}C[/tex] y [tex]T_{f} = 100\,^{\circ}C[/tex], entonces la longitud final de la varilla es:

[tex]l_{f} = (1.20\,m)\cdot \left[1 + \left(1.4\times 10^{-5}\,^{\circ}C^{-1}\right)\cdot (100\,^{\circ}C-18\,^{\circ}C)\right][/tex]

[tex]l_{f} = 1.201\,m[/tex]

La longitud de la varilla de cobre es de 1.201 metros a una temperatura de 100 °C.

Answer:  

GIVEN:

v₀=0ms⁻¹

a= 8.1ms⁻²

t= 19.4s

REQUIRE:

d=?

CALCULATUION:

as we know,

d=v₀t+1/2at²

by putting values

d=0ms⁻¹×19.4s+1/2×8.1ms⁻²×(19.4s)²

d=0m+1/2×8.1ms⁻²×376.36s²

d=1/2×3048.516m

d=1524.258m

d≈1524m

The solubility of the sample will decrease

Answer:

a. X: Slower in gases than liquids Y: Faster in solids than gases Z: Velocity depends on medium.

Explanation:

Speed of sound is fastest in solids. Sound waves travel more quickly in solid, than of liquid and gases. Sound waves travel most slowest in gases. Speed of sound varies significantly and it depends upon medium it is travelling through.  In more rigid medium sounds velocity will be faster.

Answer:

The diagram assigned B

explanation:

Check the direction of the two vectors, their resultant must be in the same direction.

Answer:

Explanation:

Acceleration is equal to the change in velocity over the change in time, or

[tex]a=\frac{v_f-v_i}{t}[/tex] where the change in velocity is final velocity minus initial velocity. Filling in:

[tex]3.0=\frac{v_f-(-3.0)}{6.0}[/tex] Note that I made the backward velocity negative so the forward velocity in our answer will be positive.

Simplifying that gives us:

[tex]3.0=\frac{v_f+3.0}{6.0}[/tex] and then isolating the final velocity, our unknown:

3.0(6.0) = v + 3.0 and

3.0(6.0) - 3.0 = v and

18 - 3.0 = v so

15 m/s = v and because this answer is positive, that means that the car is no longer rolling backwards (which was negative) but is now moving forward.

At the ball's highest point, it has no vertical velocity, so the 6 m/s is purely horizontal. A projectile's horizontal velocity does not change, which means the ball was initially thrown with speed v such that

v cos(53°) = 6 m/s   ==>   v = (6 m/s) sec(53°) ≈ 9.97 m/s

The player shoots the ball from a height of 2.0 m, so that the ball's horizontal and vertical positions, respectively x and y, at time t are

x = (9.97 m/s) cos(53°) t = (6 m/s) t

y = 2.0 m + (9.97 m/s) sin(53°) t - 1/2 gt ²

Find the times t for which the ball reaches a height of 3.00 m:

3.00 m = 2.0 m + (9.97 m/s) sin(53°) t - 1/2 gt ²

==>   t ≈ 0.137 s   or   t ≈ 1.49 s

The second time is the one we care about, because it's the one for which the ball would be falling into the basket.

Now find the distance x traveled by the ball after this time:

x = (6 m/s) (1.49 s) ≈ 8.93 m

Answer:

I think it's option D

Explanation:

I think it's option D but not so sure

Answer:

ΔP = 1875 Pa,   P₂ = P₁ - 1875

Explanation:

Let's use Bernoulli's equation, with the subscript 1 for the widest Mars and the subscript 2 for the narrowest part, suppose that the pipe is horizontal

          P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

          P₁ -P₂ = ½ ρ (v₂² - v₁²)

suppose the fluid is water

          P₁ - P₂ = ½ 1000 (2² - 0.5²)

         ΔP = 1875 Pa

this is the pressure difference between the two sections

the pressure in the narrowest section is

           P₂ = P₁ - 1875

Answer:

Explanation:

We need to find the x-components of each of these vectors and then add them together, then we need to find the y-components of these vectors and then add them together. Let's get to that point first. That's hard enough for step 1, dontcha think?

The x-components are found by multiplying the magnitude of the vectors by the cosine of their respective angles, while the y components are found by multiplying the magnitude of the vectors by the sine of their respective angles.

Let's do the x-components for all the vectors first, so we get the x-component of the resultant vector:

[tex]F_{1x}=12 cos0[/tex] and

[tex]F_{1x}=12[/tex]

[tex]F_{2x}=9cos90[/tex] and

[tex]F_{2x}=0[/tex]

[tex]F_{3x}=15 cos126.87[/tex] and

[tex]F_{3x}=-9.0[/tex]  (the angle of 126.87 is found by subtracting the 53.13 from 180, since angles are to be measured from the positive axis in a counterclockwise fashion).

That means that the x-component of the resultant vector, R, is 3.0

Now for the y-components:

[tex]F_{1y}=12sin0[/tex] and

[tex]F_{1y}=0[/tex]

[tex]F_{2y}=9sin90[/tex] and

[tex]F_{2y}=9[/tex]

[tex]F_{3y}=15sin126.87[/tex] and

[tex]F_{3y}=12[/tex]

That means that the y-component of the resultant vector, R, is 21.

Put them together in this way to find the resultant magnitude:

[tex]R_{mag}=\sqrt{(3.0)^2+(21)^2}[/tex] which gives us

[tex]R_{mag}=21[/tex] and now for the angle. Since both the x and y components of the resultant vector are positive, our angle will be where the x and y values are both positive in the x/y coordinate plane, which is Q1.

The angle, then:

[tex]tan^{-1}(\frac{21}{3.0})=82[/tex] degrees, and since we are QI, we do not add anything to this angle to maintain its accuracy.

To sum up: The resultant vector has a magnitude of 21 N at 82°

KE = (0.5) m v²

given that : v = speed of the bottle in each case = 4 m/s when m = 0.125 kg

KE = (0.5) m v² = (0.5) (0.125) (4)² = 1 J

when m = 0.250 kg KE = (0.5) m v² = (0.5) (0.250) (4)² = 2 J

when m = 0.375 kg KE = (0.5) m v² = (0.5) (0.375) (4)² = 3 J

when m = 0.0.500 kg KE = (0.5) m v² = (0.5) (0.500) (4)² = 4 J

Answer:

The position is 8.18cm from the mirror.

Nature is b=virtual

Size is 1.82cm

Explanation:

Note that for a convex mirror, the image distance and the focal length are negative;

Given

Object height H0 = 4cm

object distance u = 18cm

Radius of curvature R = 30cm

Since f = R/2

f = 30/2

f = -15cm

Recall that:

[tex]\frac{1}{f} =\frac{1}{u}+ \frac{1}{v}\\\frac{1}{-15}=\frac{1}{18}+\frac{1}{v} \\\frac{1}{v} =\frac{1}{-15} -\frac{1}{18}\\ \frac{1}{v} = \frac{-18-15}{270}\\\frac{1}{v} = \frac{-33}{270}\\v=\frac{-270}{33}\\v=-8.18cm[/tex]

Since the image distance is negative, this shows that the image is a virtual image.

To get the size:

[tex]\frac{H_1}{H_0}=\frac{v}{u}\\\frac{H_1}{4}=\frac{8.18}{18}\\18H_i=32.72\\H_i=\frac{32.72}{18}\\H_i= 1.82cm[/tex]