Which one is smaller a hectogram, dekagram, kilogram, centigram, or milligrams?

Paco was driving his scooter west with an initial velocity of 4 m/s. he accelerates at 0.5 m/s2 for 30 seconds. what is his fina

sleet_krkn [62]

The final velocity will be 19 m/sec.Option C is correct.

<h3>What is velocity?</h3>

The change of displacement with respect to time is defined as the velocity.

Velocity is a vector quantity. it is a time-based component. Velocity at any angle is resolved to get its component of x and y-direction.

From Newton's first equation of motion;

$\rm v = u+at \\\\ v= 4+ 0.5 \times 30 \\\\ v= 19 \ m/sec$

The final velocity will be 19 m/sec

Hence, option C is correct.

To learn more about the velocity refer to the link ;

brainly.com/question/862972

#SPJ1

4 0

2 years ago

A charge q = 3 × 10-6 C of mass m = 2 × 10-6 kg, and speed v = 5 × 106 m/s enters a uniform magnetic field. The mass experiences

NeX [460]

Answer:

Magnetic field, B = 0.004 mT

Explanation:

It is given that,

Charge, $q=3\times 10^{-6}\ C$

Mass of charge particle, $m=2\times 10^{-6}\ C$

Speed, $v=5\times 10^{6}\ m/s$

Acceleration, $a=3\times 10^{4}\ m/s^2$

We need to find the minimum magnetic field that would produce such an acceleration. So,

$ma=qvB\ sin\theta$

For minimum magnetic field,

$ma=qvB$

$B=\dfrac{ma}{qv}$

$B=\dfrac{2\times 10^{-6}\ C\times 3\times 10^{4}\ m/s^2}{3\times 10^{-6}\ C\times 5\times 10^{6}\ m/s}$

B = 0.004 T

or

B = 4 mT

So, the magnetic field produce such an acceleration at 4 mT. Hence, this is the required solution.

4 0

3 years ago

Which two actions would weaken an electromagnet?

Semenov [28]

Answer:

Decreasing the number of warps of wire

Replacing the steel core of the magnet with a plastic core

Explanation:

But I’m not exactly sure about your answers these are the two answers I had

7 0

3 years ago

Will give 45 points to who ever solves this

Lubov Fominskaja [6]

Answer:

1. 230 kg...

W = m * g

W= 230 kg * 9.81 m/s^2

W= 2256,3 N

m= 230kg , W = 2256,3 N , g= 9.81 m/s^2

2. 887 N

W= m * g

887 N = m * 9.81 m/s^2

m= 90,42 kg

m= 90,42 kg, W = 887 N , g= 9.81 m/s^2

3. 420 kg

W= m * a

w= 420 kg * 9.81 m/s^2

w= 4120,2 N

m= 420 kg , W = 4120,2 N , g= 9.81 m/s^2

4. Determine the gravity on Pluto where a 15 kg object weighs 55.5N.

w = m * g

55.5 N = 15 kg * g

a= 3,7 m/s^2

m = 15 kg , W= 55.5 N , g= 3,7 m/s^2

7 0

2 years ago

Read 2 more answers

A river flows due south with a speed of 2.0 m/s .You steer a motorboat across the river; your velocity relative to the water is

mihalych1998 [28]

Answer:

a) $v_m =\sqrt{v^2_x + v^2_y} = \sqrt{(2m/s)^2 +(4.8 m/s)^2}= 5.2 m/s$

b) $\theta= tan^{-1} \frac{v_y}{v_x} = tan^{-1} (\frac{2}{4.8})= 22.62$ degrees and on this case to the South of the East.

c) $t= \frac{w}{v_m}= \frac{600m}{4.8 m/s}= 125 s$

d) $Y = 2 m/s * 125 s = 250m$

So it would be 250 to the South

Explanation:

Part a

For this case the figure attached shows the illustration for the problem.

We know that $v_y = 2 m/s$ represent the velocity of the river to the south.

We have the velocity of the motorboard relative to the water and on this case is $V_x= 4.8 m/s$

And we want to find the velocity of the motord board relative to the Earth $v_m$

And we can find this velocity from the Pythagorean Theorem.

$v_m =\sqrt{v^2_x + v^2_y} = \sqrt{(2m/s)^2 +(4.8 m/s)^2}= 5.2 m/s$

Part b

We can find the direction with the following formula:

$\theta= tan^{-1} \frac{v_y}{v_x} = tan^{-1} (\frac{2}{4.8})= 22.62$ degrees and on this case to the South of the East.

Part c

For this case we can use the following definition

$D = Vt$

The distance would be D = w = 600 m and the velocity V = 4.8m/s and if we solve for t we got:

$t= \frac{w}{v_m}= \frac{600m}{4.8 m/s}= 125 s$

Part d

For this case we can use the same definition but now using the y compnent we have:

$Y = v_y t$

And replacing we got:

$Y = 2 m/s * 125 s = 250m$

So it would be 250 to the South

7 0

3 years ago