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pshichka [43]
3 years ago
15

This is my first question don’t really know how to use this app yet lol but somebody answer it for me pls!! Seded the corredare

Physics
1 answer:
miskamm [114]3 years ago
7 0
3x6



(This just for extras )
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I was driving along at 20 m/s, trying to change a CD and not watching where I was going. When I looked up, I found myself 45 m f
cricket20 [7]

Answer:

a=4.44\frac{m}{s^2}

Explanation:

First we have to find the time required for train to travel 60 meters and impact the car, this is an uniform linear motion:

t=\frac{d}{v}\\\\t=\frac{60m}{30\frac{m}{s}}=2s

The reaction time of the driver before starting to accelerate was 0.50 seconds. So, remaining time for driver is 1.5 seconds.

Now, we have to calculate the distance traveled for the driver in this 0.5 seconds before he start to accelerate. Again, is an uniform linear motion:

d=vt\\d=20\frac{m}{s}(0.5s)=10m

The driver cover 10 meters in this 0.5 seconds. So, the remaining distance to be cover in 1.5 seconds by the driver are 35 meters. We calculate the minimum acceleration required by the car in order to cross the tracks before the train arrive, Since this is an uniformly accelerated motion, we use the following equation:

d=v_0t+\frac{1}{2}at^2\\a=\frac{2(d-v_0t)}{t^2}\\a=\frac{2(35m-20\frac{m}{s}*1.5s}{(1.5s)^2}\\a=4.44\frac{m}{s^2}

7 0
3 years ago
A student drops a rock from a bridge to the water 12.7 m below. with what speed does the rock strike the water? the acceleration
kolezko [41]

The rock strike the water with the speed of 15.78 m/sec.

The speed by which rock hit the water is calculated by the formula

v=\sqrt{2gh}

v=\sqrt{2*9.8*12.7}

v=15.78 m/sec

Hence, the rock strike the water with the speed of 15.78 m/sec.

7 0
3 years ago
n deep space, sphere A of mass 47 kg is located at the origin of an x axis and sphere B of mass 110 kg is located on the axis at
Volgvan

Answer:

a)-1.014x 10^{-7J

b)3.296 x  10^{-7J

Explanation:

For Sphere A:

mass 'Ma'= 47kg

xa= 0

For sphere B:

mass 'Mb'= 110kg

xb=3.4m

a)the gravitational potential energy is given by

U_{i = -GMaMb/ d

U_{i= - 6.67 x 10^{-11} x 47 x 110/ 3.4 => -1.014x 10^{-7J

b) at d= 0.8m (3.4-2.6) and U_{i=-1.014x 10^{-7J

The sum of potential and kinetic energies must be conserved as the energy is conserved.

K_{i + U_{i= K_{f + U_{f

As sphere starts from rest and sphere A is fixed at its place, therefore K_{i is zero

U_{i= K_{f + U_{f

The final potential energy is

U_{f= - GMaMb/d

Solving for 'K_{f '

K_{f = U_{i + GMaMb/d => -1.014x 10^{-7 + 6.67 x 10^{-11} x 47 x 110/ 0.8

K_{f = 3.296 x  10^{-7J

6 0
3 years ago
Is it possible for an object that has a constant negative acceleration to change the direction in which it is moving? Explain wh
lozanna [386]
Yes! I think there are two ways you could go with this answer: 1) Acceleration is the change in velocity over time, it can be negative or positive. If you have an object that is already moving forwards in a straight line and give it a constant negative acceleration, it will slow down and then start going in reverse. 2)Velocity is a vector, meaning it has both magnitude and direction. In the example above, the acceleration is due to a change in magnitude, or speed (from +ve to -ve) but not a change in direction. Something that has constant speed but is changing direction is also accelerating (like something that is orbiting). You could use the earth as an example, which is constantly accelerating due to moving in a circle around the sun. At any time in the year you can say that in half a year's time the earth's direction will be reversed.
4 0
3 years ago
A long straight wire carries a conventional current of 0.7 A. What is the approximate magnitude of the magnetic field at a locat
vfiekz [6]

Answer:

2.64 x 10⁻⁶T

Explanation:

The magnitude of the magnetic field produced by a long straight wire carrying current is given by Biot-Savart law as follows: "The magnetic field strength is directly proportional to the current on the wire and inversely proportional to the distance from the wire". This can be written mathematically as;

B = (μ₀ I) / (2π r)                ----------------(i)

B is magnetic field

I is current through the wire

r is the distance from the wire

μ₀  is the magnetic constant = 4π x 10⁻⁷Hm⁻¹

From the question;

I = 0.7A

r = 0.053m

Substitute these values into equation (i) as follows;

B =  (4π x 10⁻⁷ x 0.7) / (2π x 0.053)

B = 2.64 x 10⁻⁶T

Therefore the approximate magnitude of the magnetic field at that location is 2.64 x 10⁻⁶T

5 0
4 years ago
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