Ask question

Ask question

All categories

3 years ago

14

A car is driving at a velocity of 24 m/s.if it brakes can supply an acceleration of -5.0 m/s^2, how much time will be required t

o bring the car to a stop.

1 answer:

UkoKoshka [18]3 years ago

4 0

Answer:The time required can be calculated by the following formula;

Explanation:

You might be interested in

How much power does it take to lift 30.0 N 10.0 m high in 5.00 s?

Rudik [331]

Power = work/time = (Force times distance)/time

= (30N *10.0m)/5.00s = 300/5 = 60 Watts

7 0

3 years ago

Read 2 more answers

Does sunlight really take 8 minutes to reach your eyes?

Brums [2.3K]

It takes sunlight 8 minutes to reach earth , so yes

3 0

3 years ago

Read 2 more answers

What is the approximate amount of thrust you need to apply to the lander to keep it's velocity roughly constant

earnstyle [38]

The approximate amount of thrust(force) you need to apply to the lander to

keep its velocity roughly constant is zero.

<h3>What is Newton's second law of motion?</h3>

Newton's second law of motion states that the acceleration the force acting

on the object is directly proportional to its rate of change of momentum.

F = m a

If the object is moving with uniform velocity, it simply means that the

acceleration is zero, and the corresponding force will also be zero.

Read more about Constant velocity here brainly.com/question/3052539

4 0

2 years ago

An electron follows a helical path in a uniform magnetic field of magnitude 0.340 T. The pitch of the path is 6.00 µm, and the m

dedylja [7]

Answer:

The electron's speed is 34007.35 m/s

Explanation:

It is given that,

Magnetic field, B = 0.34 T

Magnetic force on the electron, $F=1.85\times 10^{-15}\ N$

The electron follows a helical path. We have to find the speed of an electron. The formula for magnetic force is given by :

$F=B\times q\times v$

q = charge on an electron, $q=1.6\times 10^{-19}\ C$

v = velocity of an electron

$v=\dfrac{F}{Bq}$

$v=\dfrac{1.85\times 10^{-15}\ N}{0.34\ T\times 1.6\times 10^{-19}\ C}$

v = 34007.35 m/s

Hence, this is the required solution.

4 0

3 years ago

Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resist

Leno4ka [110]

Answer:

time=4s

Explanation:

we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion

$i(t)=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$ , where $i$ max= $\frac{E}{R}$

Given that, at i(2)= $\frac{imax}{2} =\frac{E}{2R}$

⇒ $\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})$

⇒ $\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}$

⇒ $\frac{1}{2}=e^\frac{-2}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}$

⇒ $log(2)=\frac{2}{\frac{L}{R}}$

⇒ $\frac{L}{R}=\frac{2}{log2}$

Now substitute $i(t)=\frac{3}{4}imax=\frac{3E}{4R}$

⇒ $\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$

⇒ $\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}$

⇒ $\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}$

⇒ $log(4)=\frac{t}{\frac{L}{R}}$

⇒ $t=log4\frac{L}{R}$

now subs. $\frac{L}{R}=\frac{2}{log2}$

⇒ $t=log4\frac{2}{log2}$

also $log4=log2^{2}=2log2$

⇒ $t=2log2\frac{2}{log2}$

⇒ $t=4$

5 0

3 years ago