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Ilya [14]
3 years ago
6

A girl, moving at 8.8 m/s on in-line skates, is overtaking a boy moving at 4.3 m/s as they both skate on a straight path. The bo

y tosses a ball backward toward the girl, giving it speed 11.4 m/s relative to him. What is the speed of the ball relative to the girl, who catches it?
Physics
1 answer:
BabaBlast [244]3 years ago
6 0

Answer:16.5 m/s

Explanation:

Given

Velocity of boy(v_b)=4.3 m/s

velocity of girl(v_g)=8.8 m/s

velocity of ball relative to boy(v_{Bb})=11.4 m/s backward

Now We have to find velocity of ball relative to girl

Representing it in vector form

v_b=4.3\hat{i}

v_g=8.8\hat{i}

v_{Bb}=-11.4\hat{i}

and

v_{Bb}=v_B-v_b

v_B=v_{Bb}+v_b

v_B=-11.4\hat{i}+4.3\hat{i}=-7.1\hat{i}

therefore velocity of ball relative to girl

v_{Bg}=v_B-v_g=-7.1\hat{i}-8.8\hat{i}=-16.5\hat{i}

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What Kinetic energy is exactly equal to Gravitational Potential Energy why is height halfway between the maximum height?
prohojiy [21]

Explanation:

Let us calculate the work done in lifting an object of mass m through a height h, such as in Figure 1. If the object is lifted straight up at constant speed, then the force needed to lift it is equal to its weight mg. The work done on the mass is then W = Fd = mgh. We define this to be the gravitational potential energy (PEg) put into (or gained by) the object-Earth system. This energy is associated with the state of separation between two objects that attract each other by the gravitational force

Potential energy is a property of a system rather than of a single object—due to its physical position. An object’s gravitational potential is due to its position relative to the surroundings within the Earth-object system. The force applied to the object is an external force, from outside the system. When it does positive work it increases the gravitational potential energy of the system. Because gravitational potential energy depends on relative position, we need a reference level at which to set the potential energy equal to 0. We usually choose this point to be Earth’s surface, but this point is arbitrary; what is important is the difference in gravitational potential energy, because this difference is what relates to the work done. The difference in gravitational potential energy of an object (in the Earth-object system) between two rungs of a ladder will be the same for the first two rungs as for the last two rungs.

5 0
2 years ago
In what direction must a force be applied so that the forces on the 1 kg object are balanced
choli [55]

Answer:

towards the object

Explanation:

7 0
3 years ago
If a certain mass of mercury has a volume of 0.002m3 at a temperature of 20°C, what will be the volume at 50°C,
USPshnik [31]

Set this up as a proportion.

.002 m^3/20 degrees = x/50 degrees

solve for x

x = .005 m^3

If you found this helpful, please brainliest me!

3 0
2 years ago
Consider two points in an electric field. The potential at point 1, V1, is 33 V. The potential at point 2, V2, is 175 V. An elec
Mnenie [13.5K]

Answer:

ΔU  = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

Explanation:

Since the electric potential at point 1 is V₁ = 33 V and the electric potential at point 2 is V₂ = 175 V, when the electron is accelerated from point 1 to point 2, there is a change in electric potential ΔV which is given by ΔV = V₂ - V₁.

Substituting the values of the variables into the equation, we have

ΔV = V₂ - V₁.

ΔV = 175 V - 33 V.

ΔV = 142 V

The change in electric potential energy ΔU = eΔV = e(V₂ - V₁) where e = electron charge = -1.602 × 10⁻¹⁹ C and ΔV = electric potential change from point 1 to point 2 = 142 V.

So, substituting the values of the variables into the equation, we have

ΔU = eΔV

ΔU = eΔV

ΔU = -1.602 × 10⁻¹⁹ C × 142 V

ΔU = -227.484 × 10⁻¹⁹ J

ΔU = -2.27484 × 10⁻²¹ J

ΔU ≅ -2.275 × 10⁻²¹ J

So, the required equation for the electric potential energy change is

ΔU  = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

5 0
2 years ago
What is the voltage across a semiconductor bar if the current through it is 0.17 A? The electron concentration in the bar is 2.7
Anastaziya [24]

Answer:

The voltage across a semiconductor bar is 0.068 V.

Explanation:

Given that,

Current = 0.17 A

Electron concentration n= 2.7\times10^{18}\ cm^{-3}

Electron mobility \mu=1000 cm^2/Vs

Length = 0.1 mm

Area = 500 μm²

We need to calculate the resistivity

Using formula of resistivity

\sigma=n\times q\times \mu

\rho=\dfrac{1}{\sigma}

Put the value into the formula

\rho=\dfrac{1}{2.7\times10^{18}\times10^{6}\times1.6\times10^{-19}\times1000\times10^{-4}}

\rho=2\ \mu \Omega m

We need to calculate the resistance

Using formula of resistance

R=\dfrac{\rho l}{A}

R=\dfrac{2\times10^{-6}\times0.1\times10^{-3}}{500\times(10^{-6})^2}

R=0.4\ \Omega

We need to calculate the voltage

Using formula of voltage

V= IR

Put the value into the formula

V=0.17\times0.4

V=0.068\ V

Hence, The voltage across a semiconductor bar is 0.068 V.

6 0
3 years ago
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