If the girl skater has a mass of 30 kg and moves backward at 5 m/s, what is the velocity or the boy skater
his mass is 50 kg?

Answer: 84 %

Explanation:

Answer:

By talking about something that interesto you’

sorry if wrong

Explanation:

Answer; they are larger and made of rocky material

Answer:

The box will not move from its position.

Explanation:

First, we will calculate the static frictional force that is stopping the box to move from its position:

[tex]f = \mu R = \mu W=\mu mg[/tex]

where,

f = static frictional force = ?

μ = coefficient of static friction = 0.3

m = mass of box = 4 kg

g = acceleration due to gravity = 9.81 m/s²

Therefore,

[tex]f = (0.3)(4\ kg)(9.81\ m/s^2)\\f=11.77\ N[/tex]

Since the frictional force (11.77 N) is greater than the applied force (10 N).

Therefore, the box will not move from its position.

Answer:

[tex]0.0035\ \text{m}[/tex]

Explanation:

y = Distance from the center point

d = Separation between slits = 0.3 mm

D = Distance between slit and screen = 1.5 m

[tex]\lambda[/tex] = Wavelength = 700 nm

m = Order = 1

We have the relation

[tex]d\dfrac{y}{D}=m\lambda\\\Rightarrow y=\dfrac{Dm\lambda}{d}\\\Rightarrow y=\dfrac{1.5\times 1\times 700\times 10^{-9}}{0.3\times 10^{-3}}\\\Rightarrow y=0.0035\ \text{m}[/tex]

The distance from the screen at which the first bright fringe beyond the center fringe appear is [tex]0.0035\ \text{m}[/tex].

Answer:

d= 23.25 m

Explanation:

       [tex]E_{o} = K_{transo} + K_{roto} (1)[/tex]

       [tex]E_{f} = m*g*h (2)[/tex]

       [tex]h = d* sin 37 (3)[/tex]

       [tex]E_{f} = m*g* d * sin 37 (4)[/tex]

       [tex]v = \omega * R (5)[/tex]

       [tex]K_{rot} = \frac{1}{2}* I * \omega^{2} (6)[/tex]

       [tex]E_{o} = K_{transo} + K_{roto} = \frac{1}{2}* m* v^{2} +(\frac{1}{2}* \frac{1}{2}) *m*r^{2}*(\frac{v}{r}) ^{2} = \\ \frac{3}{4} * m * v^{2} (7)[/tex]

       [tex]d =\frac{3}{4} * \frac{v^{2}}{g*sin37} = \frac{3}{4}*\frac{(\omega*r)^{2}}{g*sin 37} (8)[/tex]

       [tex]d =\frac{3}{4}*\frac{(\omega*r)^{2}}{g*sin 37} = \frac{3}{4} *\frac{(12.0536rad/sec*1.12m)^{2}}{9.8 m/s2*0.601} = 23. 25 m.[/tex]

Answer:

C

Explanation:

C

Answer:

get egg and try to make in not crack when it falls by exerting the momentum of the fall into something other than the egg ex. make a box full of bubble wrap and put your egg in it

Explanation:

Answer:

1) m = 0.3137 kg

2a)T_f = -181.7°C

2b) T_f = -1176.97°C

Explanation:

1) We are given;

Length; l = 4.30 cm = 0.043 m

Width; w = 14.0 cm = 0.014 m

height; h = 27.0 cm = 0.027 m

density of gold; ρ = 19.3 × 10³ kg/m³

Formula for the density is known as;

ρ = mass/volume

Thus;

m =ρV

m = 19.3 × 10³ × (lwh)

m = 19.3 × 10³ × (0.043 × 0.014 × 0.027)

m = 0.3137 kg

2a) We are given;

Diameter of brass; L_br = 10 cm

Diameter of aluminum; L_al = 10.01 cm

Now, to some for change in temperature we will use the formula;

L_f = L_i + αL_i(Δt)

Where α is coefficient of expansion.

Now, for the ring to be removed from the rod, the final diameter of the brass has to be same as the aluminium.

Thus;

L_f(brass) = L_f(aluminium)

From table attached, α_brass ≈ 19 × 10^(-6) /°C

Also, α_aluminium ≈ 24 × 10^(-6) /°C

Thus;

L_f(brass) = 10 + (19 × 10^(-6) × 10 × (Δt))

Similarly,

L_f(aluminium) = 10.01 + (24 × 10^(-6) × 10.01 × (Δt))

Since L_f(brass) = L_f(aluminium), then;

10 + (19 × 10^(-6) × 10 × (Δt)) = 10.01 + (24 × 10^(-6) × 10.01 × (Δt))

Rearranging, we have;

10.01 - 10 = (19 × 10^(-6) × 10 × (Δt)) - (24 × 10^(-6) × 10.01 × (Δt))

0.01 = Δt(-50.24 × 10^(-6))

Δt = 0.01/(-50.24 × 10^(-6))

Δt ≈ -199°C

Thus, temperature at which the combination must be cooled to separate the two metals is;

T_f = T_i + Δt

T_f = 17.3 + (-199)

T_f = -181.7°C

2b) Diameter of aluminum is now;

L_al = 10.06 cm

Thus;

10.06 - 10 = (19 × 10^(-6) × 10 × (Δt)) - (24 × 10^(-6) × 10.01 × (Δt))

0.06 = Δt(-50.24 × 10^(-6))

Δt = 0.06/(-50.24 × 10^(-6))

Δt = -1194.27°C

T_f = 17.3 + (-1194.27)

T_f = -1176.97°C

im confused hold on imma send you a link to the answerExplanation:

Answer:

v = -17.94 cm

Explanation:

Given that,

The candle is placed at a distance of 50 cm, u = -50 cm

The focal length of the diverging lens, f = -28 cm (negative in case of a concave lens)

We need to find the image distance. We know that the lens formula is as follows:

[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{(-28)}+\dfrac{1}{(-50)}\\\\v=-17.94\ cm[/tex]

So, the image distance is equal to 17.94 cm.

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I hope this helped!+*

Answer:

Gravity

Explanation:

Gravity pulls things down to earth and it is a force

Answer:

C

Explanation:

The men bow to the women and the women curtsy to the men.

(took on test and got it right)

Answer:

[tex]0.0018833\ \text{m/s}[/tex]

Explanation:

[tex]d[/tex] = Distance of Andromeda Galaxy from Earth = [tex]2.54\times 10^7\ \text{ly}[/tex]

[tex]t[/tex] = Time taken = [tex]90\ \text{years}[/tex]

[tex]c[/tex] = Speed of light = [tex]3\times 10^8\ \text{m/s}[/tex]

We have the relation

[tex]t=t_o\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow 90=2.54\times 10^7\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow \dfrac{90^2}{(2.54\times 10^7)^2}=1-\dfrac{v^2}{c^2}\\\Rightarrow 1-\dfrac{90^2}{(2.54\times 10^7)^2}=\dfrac{v^2}{c^2}\\\Rightarrow v=c\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}}[/tex]

[tex]c-v=c(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=3\times 10^8(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=0.0018833\ \text{m/s}[/tex]

The required answer is [tex]0.0018833\ \text{m/s}[/tex].

Answer:

the answer is false.

Explanation:

i took the test and it is false trust me!!!!!!!!!

Answer:

False

Explanation:

No, it is not true that energy of the electrons ejected should have been proportional to the intensity of the light. Perhaps the kinetic energy of the ejected electrons is independent of the intensity of the incident radiation. This is the very fact that classical theory of electromagnetism fails to explain in photoelectric effect. The kinetic energy of the electrons remains constant even if the amplitude of the incident light is increased.

Answer:

Amplitude

Explanation:

The loudness of a sound depends on the amplitude of vibration producing the sound

Answer:

12294.31 m/s

Explanation:

Momentum = (mass)(velocity)

Momentum before = Momentum after

(momentum of bullet)+(momentum of block)=(momentum of bullet and block)

0.035v+50(0)=(0.035+50)(8.6)

0.035v=430.301

v=12294.3142857m/s

Answer:

B : using a spring scale and measuring the weight of the car

Explanation:

Answer:

Waves carry energy from one place to another.

Explanation:

Because waves carry energy, some waves are used for communication, eg radio and television waves and mobile telephone signals.

Answer:

It will go up to 93.75 m before it is moving at 20 m/s

Explanation:

As we know that

[tex]v^2 - u^2 = 2aS[/tex]

here v is the final speed i.e 20 m/s

u is the initial speed i.e 5 m/s

a is the acceleration due to gravity i.e 2 m/s^2

Substituting the given values in above equation, we get -

[tex]20^2 - 5^2 = 2*2*S\\S = 93.75[/tex]meters

Answer:

Time taken by Mr. smith = 4.80 hour (Approx.)

Explanation:

Given:

Distance travel by Mr. smith = 523 kilometer

Average speed of Mr. smith = 109 km/hr

Find;

Time taken by Mr. smith

Computation:

Time taken = Distance cover / Speed

Time taken by Mr. smith = Distance travel by Mr. smith / Average speed of Mr.

smith

Time taken by Mr. smith = 523 / 109

Time taken by Mr. smith = 4.798 hr

Time taken by Mr. smith = 4.80 hour (Approx.)

Answer:

please give me brainlist and follow

Explanation:

4 degrees C turns out to be the temperature at which liquid water has the highest density. If you heat it or cool it, it will expand. ... Ice floats on top of lakes, preventing evaporation (and convection in the frozen layer), and lakes stay liquid underneath, allowing fish and other life to survive.

Answer:

the car movves briefly as you ran, however, it stops again after you ran in to the wall

Explanation:

Our net total linear momentum was zero at the time both the train and the boy was at rest. As there is no pressure, the train and boy system's linear momentum can be conserved provided no external forces are working on it that might shift its momentum.

If the boy runs inside of the train with a velocity V1 in the forward direction, the train would have a velocity V2 in the reverse direction to V1 to conserve the system's momentum, resulting in Final momentum.

i.e

[tex]m \times V_1 + M \times V_2 = 0[/tex]  --- (1)

here;

m = mass of the boy

M = mass of the train

Thus;

[tex]m \times V_1 =- M \times V_2[/tex] --- (2)

As the boy crashes into the train's wall, a pair of equal and opposing force F intervene on both the train and the boy. This force F causes the boy traveling with momentum m× V1 to come to a halt; its momentum remains zero. As the train moves with about the same momentum as the boy as seen in (2) and is subjected to the same force F, its momentum will be diminished to zero, and it will come to a halt.

Answer:

What is the chemical fórmula For the molecule modeled?

Explanation:

C6H12O2

Answer:

a)   y = 3.73 m,  b)  t = 1.74 s,  c)  R = 40.99 m,

d) vₓ  = 23.49  m/s,   v_y = -8.5 m / s

Explanation:

This is a projectile launching exercise, we start by breaking down the initial velocity

          sin θ = v_{oy} / v₀

          cos θ = v₀ₓ / v₀

          v_{oy} = v₀ sin θ

          v₀ₓ = v₀ cos θ

          v_{oy} = 25 sin 20 = 8.55 m / s

          v₀ₓ = 25 cos 20 = 23.49 m / s

a) when the egg reaches the maximum height its vertical speed is zero

          v_y² = v_{oy}² - 2 g y

          0 = v_[oy}² - 2g y

           y = v_{oy}² / 2g

          y = [tex]\frac{8.55^2}{2 \ 9.8 }[/tex]

          y = 3.73 m

b) flight time

          y = v_{oy} t - ½ g t²

the time of flight occurs when the body reaches the ground y = 0

          0 = (v_{oy} - ½ g t) t

The results are

          t₁ = 0s        this time is for using the body star

          v_{oy} - ½ g t = 0

           t = [tex]\frac{2v_{oy}^2}{g}[/tex]

           t = 2 8.55 / 9.8

           t = 1.74 s

c) the range

           R = v₀² sin 2θ / g

           R = 25² sin (2 20) / 9.8

           R = 40.99 m

d) speed at the point of arrival

horizontal speed is constant

           vₓ = v₀ₓ = 23.49  m/s

vertical speed is

           v_y = Iv_{oy} - g t

           v_y = 8.55 - 9.8  1.74

           v_y = -8.5 m / s