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2 years ago

What is the instantaneous velocity v of the particle at t=10.0s?

1 answer:

8 0

Instantaneous velocity is defined at a specific value of time t. It can never be observed or measure. To know a velocity, one must know its distance and velocity. It is represented by the equation velocity is equal to delta distance over delta time. The problem above lacks data.

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Two identical balls are at rest and side by side at the top of a hill. You let one ball, A, start rolling down the hill. A littl

ICE Princess25 [194]

Answer:

Option b. it has the same position and the same acceleration as A

Explanation:

Let's analyze every statement:

a. it has the same position and the same velocity as A

In the instant where B passes A, they Do have the same position. Velocity however, cannot be the same because if they were, ball B would never pass ball A. So, this is false.

b. it has the same position and the same acceleration as A

As we said in the previous option, the position is the same. The acceleration is gravity for both balls, so this is true.

c. it has the same velocity and the same acceleration as A

Acceleration is the same but velocities are not, so this is false.

d. it has the same displacement and the same velocity as A

The distance they have traveled is the same, so the displacement is the same, but the velocity is not, so this is false.

e. it has the same position, displacement and velocity as A

The position and displacement is the same but not velocity, so this is false.

Only option b is true.

3 0

1 year ago

You toss a rock of mass m vertically upward. Air resistance can be neglected. The rock reaches a maximum height h above your han

VladimirAG [237]

Answer with Explanation:

We are given that

Mass of rock=m

Maximum height=h

a.At maximum height, velocity,v=0

We know that

$v^2=u^2-2gh$

$0+2gh=u^2$

$u^2=2gh$

Height,h=h/4

Again, $v'^2=u^2-2g\times \frac{h}{4}$

$v'^2=2gh-\frac{gh}{2}=\frac{4gh-gh}{2}=\frac{3gh}{2}$

$v'=\sqrt{\frac{3gh}{2}}=\sqrt{\frac{3\times 9.8 h}{2}}=3.83\sqrt h$

Where $g=9.8 m/s^2$

b.When height,h=3h/4

$v'^2=u^2-2gh$

$v'^2=2gh-2g\times \frac{3h}{4}=2gh-\frac{3gh}{2}=\frac{4gh-3gh}{2}=\frac{gh}{2}$

$v'=\sqrt{\frac{9.8h}{2}}=2.2\sqrt h$

$v'=2.2\sqrt h$

4 0

2 years ago

Read 2 more answers

A particle has a velocity of v→(t)=5.0ti^+t2j^−2.0t3k^m/s.

Makovka662 [10]

Answer:

a) $a=5 i+2t j - 6\ t^2k$

b) $a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2$

Explanation:

Given that

v(t) = 5 t i + t² j - 2 t³ k

We know that acceleration a is given as

$a=\dfrac{dv}{dt}$

$\dfrac{dv}{dt}=5 i+2t j - 6\ t^2k$

$a=5 i+2t j - 6\ t^2k$

Therefore the acceleration function a will be

$a=5 i+2t j - 6\ t^2k$

The acceleration at t = 2 s

a= 5 i + 2 x 2 j - 6 x 2² k m/s²

a=5 i + 4 j -24 k m/s²

The magnitude of the acceleration will be

$a=\sqrt{5^2+4^2+24^2}\ m/s^2$

a= 24.83 m/s²

The direction of the acceleration a is given as

$a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2$

a) $a=5 i+2t j - 6\ t^2k$

b) $a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2$

5 0

2 years ago

When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in pa

MA_775_DIABLO [31]

Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

I = Current = 0.33 A (series)

I = Current = 1.6 A (parallel)

In series

$\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2$

In parallel

$\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}$

$\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0$