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2 years ago

If a girl is running along a straight road with a uniform velocity 1.5 m /s find her acceleration

Physics

1 answer:

Phoenix [80]2 years ago

6 0

Answer:

As velocity is uniform then acceleration will be zero.

Explanation:

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Can someone solve this problem and explain to me how you got it

likoan [24]

Answer:

Explanation:

Coulomb's law states that the force of attraction or repulsion between any two charges is proportional to the product of the magnitude of charges and inversely proportional to the square of the distance between the charges

⇒ $F\alpha\frac{q1*q2}{r^{2}}$

∴ $F=k\frac{q1*q2}{r^{2}}$

where $F$ is the force of attraction or repulsion

$k$ is Coulumb's constant= $9*10^{9}Nm^{2}C^{-2}$

$q1$ and $q2$ are the magnitude of the charges

$r$ is the distance between two charges

The force between the two charges is attractive if they are of different polarity

The force between the two charges is repulsive if they are of same polarity

Question 7:

Given: $F=1.9*10^{-29}N$ , $q1=q2=1.6*10^{-19}C$

By coulomb's law,

$1.9*10^{-29}=9*10^{9}*\frac{1.6*10^{-19}*1.6*10^{-19}}{r^{2}}$

⇒ $r=3.5m$

Question 6:

Given: $q1=q2=-1.5*10^{-6}C$ , $r=0.28m$

By coulomb's law,

$F=9*10^{9}*\frac{1.5*10^{-6}*1.5*10^{-6}}{0.28^{2}}$

⇒ $F=0.26N$

3 0

2 years ago

I need help on this.

AVprozaik [17]

Answer:

A. Total mass of each team

Explanation:

The side with the larger mass will end up pulling the side with the smaller mass

6 0

3 years ago

A ball is thrown vertically upward from the edge of a bridge 22.0 m high with an initial speed of 16.0 m/s. The ball falls all t

KonstantinChe [14]

To solve this problem we will apply the concepts related to the kinematic equations of motion. We will start calculating the maximum height with the given speed, and once the total height of fall is obtained, we will proceed to calculate with the same formula and the new height, the speed of fall.

The expression to find the change in velocity and the height is,

$v_f^2-v_0^2 = -2gh$

Replacing,

$0^2-16^2 = -2(9.8)h$

$h = 13.0612m$

Thus the total height reached by the ball is

H = 22m+13.0612m

H = 35.0612m

Now calculate the velocity while dropping down from the maximum height as follows

$v_f^2-v_0^2 = 2gh$

Substituting the new height,

$v_f^2 - 0^2 = 2(9.8)(35.0612)$

$v = \sqrt{687.2}$

$v = 26.2145m/s$

3 0

3 years ago

A roadrunner is running along a straight desert road at a constant velocity of 25 m/s. If a certain coyote wants to capture the

andreyandreev [35.5K]

Answer:

t = 1.42 s and d = 35.5 m

Explanation:

Given that,

Velocity of a roadrunner is 25 m/s

A certain coyote wants to capture the roadrunner using a net dropped from an overpass that is 10 m high.

We need to find the time before the roadrunner is under the overpass and how far away from the overpass is the roadrunner when the coyote drops the net.

$d=ut+\dfrac{1}{2}at^2\\\\\text{Here, u = 0 and a = g}\\\\d=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2d}{g}} \\\\t=\sqrt{\dfrac{2\times 10}{9.8}} \\\\t=1.42\ s$

Let d is the distance traveled. So,

d = vt

d = 25 m/s × 1.42 s

d = 35.5 m

5 0

2 years ago

A force of 36 Newtons causes an object to accelerate from rest to a speed of 12 m/s in 6.0 seconds. What is the mass of the obje

VladimirAG [237]

Answer:

Explanation:

7 0

2 years ago

If a girl is running along a straight road with a uniform velocity 1.5 m /s find her acceleration ​

If a girl is running along a straight road with a uniform velocity 1.5 m /s find her acceleration