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WARRIOR [948]
3 years ago
15

If the resistance of a circuit increases, that means that the current has to

Physics
1 answer:
Gnoma [55]3 years ago
4 0

Answer:

If resistance increases current decreases.

Explanation:

  • Current is <em>inversely proportional</em> to the resistance.
  • from the relation given below, we can clearly see the relation between current and resistance;

                              V=IR

                              I ∝ 1/R

This relation shows that when resistance increases,current decreases.

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A marble runs off the edge of a table that is 1.5 m high and the marble lands 0.50 m from the base of the table. a. How much tim
Free_Kalibri [48]

Answer:

t = 0.55[sg]; v = 0.9[m/s]

Explanation:

In order to solve this problem we must establish the initial conditions with which we can work.

y = initial elevation = - 1.5 [m]

x = landing distance = 0.5 [m]

We set "y" with a negative value, as this height is below the table level.

in the following equation (vy)o is equal to zero because there is no velocity in the y component.

therefore:

y = (v_{y})_{o}*t - \frac{1}{2} *g*t^{2}\\   where:\\(v_{y})_{o}=0[m/s]\\t = time [sg]\\g = gravity = 9.81[\frac{m}{s^{2}}]\\ -1.5 = 0*t -4.905*t^{2} \\t = \sqrt{\frac{1.5}{4.905} } \\t=0.55[s]

Now we can find the initial velocity, It is important to note that the initial velocity has velocity components only in the x-axis.

(v_{x} )_{o} = \frac{x}{t} \\(v_{x} )_{o} = \frac{0.5}{0.55} \\(v_{x} )_{o} =0.9[m/s]

3 0
3 years ago
HELP ME PLS THIS IS DUE TODAY I NEED HELP!!!!
lions [1.4K]

Explanation:

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4 0
3 years ago
A 0.145-kg baseball pitched horizontally at 35.0 m/s strikes a bat and is popped straight up to a height of 40.0m. If the contac
Nookie1986 [14]

Answer:

F = 2.6 \times 10^3 N

Explanation:

Maximum height reached by the ball after being popped by the bat is given as

y = 40 m

now we know by energy conservation final speed of the ball after being hit by the bat is given as

\frac{1}{2}mv^2 = mgh

v = \sqrt{2gh}

v = \sqrt{2(9.81)(40)}

v = 28 m/s

now the change in momentum of the ball is given as

\Delta P = m(v_f - v_i)

\Delta P = 0.145(28\hat j + 35 \hat i)

now force is given as rate of change in momentum

F = \frac{\Delta P}{\Delta t}

F = \frac{0.145(28\hat j + 35 \hat i)}{2.5 \times 10^{-3}}

F = (1.62 \hat j + 2.03 \hat i)\times 10^3 N

so magnitude of the force is given as

F = \sqrt{1.62^2 + 2.03^2} \times 10^3 N

F = 2.6 \times 10^3 N

4 0
3 years ago
Someone please answer all 5 will mark brainliest on first answer!!!!
Elanso [62]

Answer:

Explanation:

Electrons play a major role in all chemical bonds. There is one type of bonding called electrovalent bonding (ionic), where an electron from one atom is transferred to another atom. You wind up creating two ions as one atom loses an electron and one gains one.

7 0
3 years ago
Please help me with the formular of this question and the solvings.​
sweet [91]

Answer:

\displaystyle T=48.86\ N

Explanation:

<u>Net Force </u>

The second Newton's law explains how to understand the dynamics of a system where several forces are acting. The forces are vectorial magnitudes which means the x and y coordinates must be treated separately. For each component, the net force must equal the mass by the acceleration, i.e.

F_{nx}=ma_x

F_{ny}=ma_y

The box with mass m=20 kg is pulled by a rope with a \theta= 30^o angle above the horizontal. It means that force (called T) has two components:

T_x=Tcos\theta

T_y=Tsin\theta

We'll assume the positive directions are to the right and upwards and that the box is being pulled to the right. There are two forces in the x-axis: The x-component of T (to the right) and the friction force (to the left). So the equilibrium equation for x is

\displaystyle T\ cos\theta -Fr=m.a

There are three forces acting in the y-axis: The component of T (upwards), the weight (downwards), and the Normal (upwards). Since there is no movement in the y-axis, the net force is zero and:

\displaystyle N+T\ sin\theta -mg=0

Rearranging:

\displaystyle N+T\ sin\theta =mg

Solving for N in the y-axis:

\displaystyle N=mg-T\ sin\theta

The friction force is given by

\displaystyle Fr=\mu.N

Replacing in the equation for the x-axis, we have

\displaystyle T\ cos\theta -\mu\ N=ma

Replacing the formula for N in the equation for the x-axis  

\displaystyle T\ cos\theta -\mu(mg-T\ sin\theta)=ma

Operating and rearranging

\displaystyle T\ cos\theta -\mu\ mg+T\ \mu\ sin\theta=ma

\displaystyle T\ (cos\theta +\mu\ sin\theta)=ma +\mu\ mg

Solving for T:

\displaystyle T=\frac{a+\mu\ g}{cos\theta +\mu\ sin\theta }\ m

Plugging in the given values:

\displaystyle T=\frac{0.4+0.2(9.8)}{cos30^o+0.2\ sin30^o }\ .20

\boxed{\displaystyle T=48.86\ N}

7 0
3 years ago
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