Ask question

Ask question

All categories

4 years ago

13

A particle traveling in a circular path of radius 300 m has an instantaneous velocity of 30 m/s and its velocity is increasing a

t a constant rate of 4 m/s2. What is the magnitude of its total acceleration at this instant?

1 answer:

hodyreva [135]4 years ago

4 0

Answer:

5 m/s2

Explanation:

The total acceleration of the circular motion is made of 2 components: centripetal acceleration and linear acceleration of 4 m/s2. They are perpendicular to each other.

The centripetal acceleration is the ratio of instant velocity squared and the radius of the circle

$a_c = \frac{v^2}{r} = \frac{30^2}{300} = \frac{900}{300} = 3 m/s^2$

So the magnitude of the total acceleration is

$a = \sqrt{a_c^2 + a_l^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 m/s^2$

You might be interested in

Which type of light wave has the highest energy? which has the least amount of energy

devlian [24]

Answer:

Red has the lowest energy and violet the highest. Beyond red and violet are many other kinds of light our human eyes can't see, much like there are sounds our ears can't hear. On one end of the electromagnetic spectrum are radio waves, which have wavelengths billions of times longer than those of visible light.

Explanation:

6 0

3 years ago

A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh

Elanso [62]

Answer:

The value is $v = 2.3359 *10^{4} \ m/s$

Explanation:

From the question we are told that

The initial speed is $u = 2.05 *10^{4} \ m/s$

Generally the total energy possessed by the space probe when on earth is mathematically represented as

$T__{E}} = KE__{i}} + KE__{e}}$

Here $KE_i$ is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

$KE_i = \frac{1}{2} * m * u^2$

=> $KE_i = \frac{1}{2} * m * (2.05 *10^{4})^2$

=> $KE_i = 2.101 *10^{8} \ \ m \ \ J$

And $KE_e$ is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

$KE_e = \frac{1}{2} * m * v_e^2$

Here $v_e$ is the escape velocity from earth which has a value $v_e = 11.2 *10^{3} \ m/s$

=> $KE_e = \frac{1}{2} * m * (11.3 *10^{3})^2$

=> $KE_e = 6.272 *10^{7} \ \ m \ \ J$

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

$KE_p = \frac{1}{2} * m * v^2$

Generally from the law energy conservation we have that

$T__{E}} = KE_p$

So

$2.101 *10^{8} m + 6.272 *10^{7} m = \frac{1}{2} * m * v^2$

=> $5.4564 *10^{8} = v^2$

=> $v = \sqrt{5.4564 *10^{8}}$

=> $v = 2.3359 *10^{4} \ m/s$

4 0

3 years ago

Your front teeth can act as a simple machine. Which machine are they?

lawyer [7]

<span>Wedges is your answer please mark brainliest </span>

4 0

3 years ago

Read 2 more answers

A long copper wire of radius 0.321 mm has a linear charge density of 0.100 μC/m. Find the electric field at a point 5.00 cm from

krek1111 [17]

Answer:

$E=35921.96N/C$

Explanation:

From the question we are told that:

Radius $r=0.321mm$

Charge Density $\mu=0.100$

Distance $d= 5.00 cm$

Generally the equation for electric field is mathematically given by

$E=\frac{mu}{2\pi E_0r}$

$E=\frac{0.100*10^{-6}}{2*3.142*8.86*10^{-12}*5*10^{-2}}$

$E=35921.96N/C$

4 0

3 years ago

A cord of mass 0.65 kg is stretched between two supports 8.0 m apart. If the tension in the cord is 120 N, how long will it take

DedPeter [7]

The time taken by the pulse to travel from one support to the other is 0.208 s.

<h3>Given:</h3>

The mass of the cord is m = 0.65 kg.

The distance between the supports is, d = 8.0 m.

The tension in the cord is T = 120 N.

The time taken by the pulse to travel from one support to the other is given as,

$v=\frac{d}{t}$

$t=\frac{d}{v}$

Here, v is the linear velocity of a pulse. Its value is,

$v=\sqrt{\frac{T d }{m} }$

$v=\sqrt{\frac{120 * 8}{0.65} }$

$v= 38.43 m/s$

Then,

$t=\frac{8}{38.43}$

$t=0.208 s$

Thus, the time taken by the pulse to travel from one support to the other is 0.208 s.

Learn more about tension here:

brainly.com/question/24994188

#SPJ4

7 0

2 years ago