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3 years ago

A car accelerates at a rate of 4 m/s in 5 s. Calculate the change of velocity PLS HELP

Physics

1 answer:

garri49 [273]3 years ago

6 0

Answer:

Explanation:

a=Δv/Δt

Δv=a*Δt

a=4 m/s2

t=5 s

Δv=4*5

Δv=20 m/s

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goldfiish [28.3K]

Answer:

Explanation:

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3 years ago

A point charge q1 = 1.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 1 m.

iris [78.8K]

To solve this problem we will apply the concepts related to the Electrostatic Force given by Coulomb's law. This force can be mathematically described as

$F = \frac{kq_1q_2}{d^2}$

Here

k = Coulomb's Constant

$q_{1,2} =$ Charge of each object

d = Distance

Our values are given as,

$q_1 = 1 \mu C$

$q_2 = 6 \mu C$

d = 1 m

$k = 9*10^9 Nm^2/C^2$

a) The electric force on charge $q_2$ is

$F_{12} = \frac{ (9*10^9 Nm^2/C^2)(1*10^{-6} C)(6*10^{-6} C)}{(1 m)^2}$

$F_{12} = 54 mN$

Force is positive i.e. repulsive

b) As the force exerted on $q_2$ will be equal to that act on $q_1$ ,

$F_{21} = F_{12}$

$F_{21} = 54 mN$

Force is positive i.e. repulsive

c) If $q_2 = -6 \mu C$ , a negative sign will be introduced into the expression above i.e.

$F_{12} = \frac{(9*10^9 Nm^2/C^2)(1*10^{-6} C)(-6*10^{-6} C)}{(1 m)^{2}}$

$F_{12} = F_{21} = -54 mN$

Force is negative i.e. attractive

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3 years ago

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nata0808 [166]

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3 years ago

Read 2 more answers

To get off a frozen lake, a 70 kg person removes his shoe of mass 0.175 kg and throws it horizontally away from the shore at a v

MakcuM [25]

Answer:

the person will be in the shore at 10.73 minutes after launch the shoe.

Explanation:

For this we will use the law of the lineal momentum.

$L_i = L_f$

Also,

L = MV

where M is de mass and V the velocity.

replacing,

$M_i V_i = M_{fp}V_{fp} + M_{fz}V_{fz}$

wher Mi y Vi are the initial mass and velocity, Mfp y Vfp are the final mass and velocity of the person and Mfz y Vfz are the final mass and velocity of the shoe.

so, we will take the direction where be launched the shoe as negative. then:

(70)(0) = (70-0.175)( $V_fp$ ) + (0.175)(-3.2m/s)

solving for $V_fp$ ,

$V_fp$ = $\frac{(3.2)(0.175)}{69.825}$

$V_fp$ = 0.008m/s

for know when the person will be in the shore we will use the rule of three as:

1 second -------------- 0.008m

t seconds-------------- 5.15m

solving for t,

t = 5.15m/0.008m

t = 643.75 seconds = 10.73 minutes

3 0

3 years ago