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3 years ago

Which two actions would weaken an electromagnet?

2 answers:

5 0

Answer: A and B (A. Replacing the steel core of the magnet with a plastic core and B. Decreasing the number of wraps of wire)

Explanation: just took the test

mario62 [17]3 years ago

3 0

Answer:

a and b

Explanation:

a pex

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In order to get unlimited answers without buying brainly plus you should.....

Papessa [141]

Answer:

What I do is answer a question every few weeks or so and I just get a bunch of answers until it starts making me watch videos again.

Explanation:

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3 years ago

The transfer of energy through varios stages in an ecological community is called

salantis [7]

It is called food chain

6 0

4 years ago

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Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant

$m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{4}}$ and $C$ is a constant

$\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0$ because $v_{(x)}$ is a constant in this derivation respect to $t$

$m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{2}}$ and $C$ is a constant

6 0

3 years ago

Two positive point charges, each of which has a charge of 2.5 × 10−9 C, are located at y = + 0.50m and y = − 0.50m. Find the mag

nadya68 [22]

Answer:

F = 147,78*10⁻⁹ [N]

Explanation:

By symmetry the Fy components of the forces acting on charge in point x = 0,7 m canceled each other, and the total force will be twice Fx ( Fx is x axis component of one of the forces .

The angle β ( angle between the line running through one of the charges in y axis and the charge in x axis) is

tan β = 0,5/0,7

tan β = 0,7142 then β = arctan 0,7142 ⇒ β = 35 ⁰

cos β = 0,81

d = √ (0,5)² + (0,7)² d1stance between charges

d = √0,25 + 0,49

d = √0,74 m

d = 0,86 m

Now Foce between two charges is:

F = K* q₁*q₂/ d² (1)

Where K = 9*10⁹ N*m²/C²

q₁ = 2,5* 10⁻⁹C

q₂ = 3,0*10⁻⁹C

d² = 0,74 m²

Plugging these values in (1)

F = 9*10⁹* 2,5* 10⁻⁹*3,0*10⁻⁹ / 0,74 [N*m²/C²]*C*C/m²

F = 91,21 * 10⁻⁹ [N]

And Fx = F*cos β

Fx = 91,21 * 10⁻⁹ *0,81

Fx =73,89*10⁻⁹ [N]

Then total force acting on charge located at x = 0,7 m is:

F = 2* Fx

F = 2*73,89*10⁻⁹ [N]

F = 147,78*10⁻⁹ [N]

8 0

3 years ago

An object initially at rest accelerates at 5 meters per second2 until it attains a speed of 30 meters per second. What distance

Tcecarenko [31]

Answer: 90m

Explanation:

Use Equation for distance:

S=a*t²/2

use eqation for acceleraton a=(Vf-Vs)/t

Vs- starting speed

Vf - final speed

a-accelaration

---------------------------------

a=5m/s²

Vs=0m/s

Vf=30m/s

use

a=(Vf-Vs)/t

to find time

t=(Vf-Vs)/a

t=30m/s/5m/s²

t=6s

Now calculate distance that object travel using:

S=(a*t²)72

s=(5m/s²*(6s)²)/2

S=90m

4 0

3 years ago

Read 2 more answers