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oksian1 [2.3K]
3 years ago
10

An object starts from rest, and accelerates at 2m/s2 for 10s. How far has it gone in that time

Physics
1 answer:
blagie [28]3 years ago
7 0

Answer:

100m

Explanation:

s = ut +  \frac{1}{2} a {t}^{2}

u=0;t=10sec;a=2m/s²

s = 0(10) +  \frac{1}{2}(2 \times  {10}^{2} )

s=10²;100m

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A police car is traveling north on a straight road at a constant 16.0 m/s. An SUV traveling north at 30.0 m/s passes the police
Nastasia [14]

Answer:

It will take 15.55s for the police car to pass the SUV

Explanation:

We first have to establish that both the police car and the SUV will travel the same distance in the same amount of time. The police car is moving at constant velocity and the SUV is experiencing a deceleration. Thus we will use two distance fromulas (for constant and accelerated motions) with the same variable for t and x:

1. x=x_{0}+vt

2. x=x_{0}+v_{0}t+\frac{at^{2}}{2}

Since both cars will travel the same distance x, we can equal both formulas and solve for t:

vt = v_{0}t+\frac{at^2}{2}\\\\   16\frac{m}{s}t =30\frac{m}{s}t-\frac{1.8\frac{m}{s^{2}} t^{2}}{2}

We simplify the fraction present and rearrange for our formula so that it equals 0:

0.9\frac{m}{s^{2}} t^{2}-14\frac{m}{s}t=0 \\\\ t(0.9\frac{m}{s^{2}}t-14\frac{m}{s})=0

In the very last step we factored a common factor t. There is two possible solutions to the equation at t=0 and:

0.9\frac{m}{s^{2}}t-14\frac{m}{s}=0 \\\\  0.9\frac{m}{s^{2}}t =14\frac{m}{s} \\\\ t =\frac{14\frac{m}{s}}{0.9\frac{m}{s^{2}}}=15.56s

What this means is that during the displacement of the police car and SUV, there will be two moments in time where they will be next to each other; at t=0 s (when the SUV passed the police car) and t=15.56s(when the police car catches up to the SUV)

8 0
3 years ago
A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the
BartSMP [9]

Answer:

r=1.14m

Explanation:

\theta is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

F_m=F_c\\qvB=F_c(1)

F_c is the centripetal force and is defined as:

F_c=m\frac{v^2}{r}

Here v is the proton's speed and r is the radius of the circular motion. Replacing this in (1) and solving for r:

qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}

Recall that 1 J is equal to 6.242*10^{12}MeV, so:

4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J

We can calculate v from the kinetic energy of the proton:

K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}

Finally, we calculate the radius of the proton path:

r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m

8 0
3 years ago
Which best explains why the shape of a liquid can change?
topjm [15]
The particles in a liquid are extremely fast, not faster than a gas, but faster than a solid. The way the particles move allow you to get volume but not area for it doesn't have a defiant shape
3 0
3 years ago
Read 2 more answers
Hey there!
eimsori [14]

The time taken for the two balls to hit each other is 8 s.

The given parameters:

  • <em>Acceleration of the rocket, a = 2 m/s²</em>
  • <em>Length of the chamber, s = 4 m</em>
  • <em>Speed of the first ball, = V1 = 0.3 m/s</em>
  • <em>Speed of the second ball, V2 = 0.2 m/s</em>

The time taken for the two balls to hit each other is calculated by applying relative velocity formula as shown below;

(V_1 - (-V_2) )t = s\\\\(V_1 + V_2) t = s\\\\(0.3 + 0.2) t = 4\\\\0.5t = 4\\\\t = \frac{4}{0.5} \\\\t = 8 \ s

Thus, the time taken for the two balls to hit each other is 8 s.

Learn more about relative velocity here: brainly.com/question/17228388

8 0
2 years ago
Read 2 more answers
Equations for physic grade 9-5
stepladder [879]
It's on your exam boards specification or just google it. the foundation paper will have the same equations as higher, but higher just has more.
6 0
3 years ago
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