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3 years ago

5

Three point charges are fixed in place in a right triangle, as shown in the figure.

1 answer:

8090 [49]3 years ago

4 0

Oh gosh oh I see it in my life face and

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A car with a mass of 710 kg is traveling at 37 km/hr. It accelerates to a speed of 120 km/hr in 12.6 seconds. What is the net fo

guajiro [1.7K]

Answer: The net force acting on the car 1,299.3 N.

Explanation:

Mass of the car = 710 kg

Initial velocity of the car of the ,u= 37 km/h= 10.27 m/s $(1km\h=\frac{5}{18} m/s)$

Final velocity of the car,v = 120 km/h = 33.33 m/s

time taken b y car = 12.6 sec

v-u=at

$33.33m/s-10.27m/s=23.06 m/s=a(12.6 sec)$

$a = 1.83 m/s^2$

$Force=mass\times acceleration$

$Force=710 kg\times 1.83 m/s^2$

$Force=1,299.3 N$

The net force acting on the car 1,299.3 N.

8 0

3 years ago

You serve a volley ball with a mass of 1.5 kg. The ball leaves your hand at 15m/s. The ball has how much kinetic energy?

pav-90 [236]

Answer:

the answer is 168.75 J

5 0

3 years ago

A 35 N force makes a 10 degree angle with the positive x-axis. What is the magnitude of the vertical component of the force?

Snowcat [4.5K]

Answer:

6.07 N

Explanation:

Given that,

Force, F = 35 N

It makes 10 degree angle with the positive x-axis.

We need to find the magnitude of the vertical component of the force. It can be given by :

$F_y=F\sin\theta\\\\=35\times \sin(10)\\\\=6.07\ N$

So, the magnitude of the vertical component of the force is 6.07 N.

5 0

3 years ago

Read 2 more answers

If the coefficient of static friction between your coffee cup and the horizontal dashboard of your car is µs = 0.800, how fast c

Cloud [144]

Answer:

Before start of slide velocity will be 14.81 m/sec

Explanation:

We have given coefficient of static friction $\mu =0.8$

Angle of inclination is equal to $\Theta =tan^{-1}\mu$

$\Theta =tan^{-1}0.8=38.65^{\circ}$

$tan{38.65^{\circ}}=0.8$

Radius is given r = 28 m

Acceleration due to gravity $g=9.8m/sec^2$

We know that $tan\Theta =\frac{v^2}{rg}$

$0.8=\frac{v^2}{28\times 9.8}$

$v^2=219.52$

$v=14.816m/sec$

So before start of slide velocity will be 14.81 m/sec

3 0

3 years ago

How to calculate the slope of position time graph?

Mashcka [7]

Answer:

The slope of a position-time graph can be calculated as:

$m=\frac{\Delta y}{\Delta x}$

where

$\Delta y$ is the increment in the y-variable

$\Delta x$ is the increment in the x-variable

We can verify that the slope of this graph is actually equal to the velocity. In fact:

$\Delta y$ corresponds to the change in position, so it is the displacement, $\Delta s$

$\Delta x$ corresponds to the change in time $\Delta t$ , so the time interval

Therefore the slope of the graph is equal to

$m=\frac{\Delta s}{\Delta t}$

which corresponds to the definition of velocity.

3 0

3 years ago