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2 years ago

9

At what constant rate of acceleration will a car starting from rest cover 18m in the first 3 seconds?

1 answer:

irga5000 [103]2 years ago

7 0

69 hehehehe hehehehe here

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An object is moving in a straight line at a constant speed. A resultant force begins to act upon the object. State the ways in w

sergiy2304 [10]

Answer:

accelerate / increase speed OR decelerate / decrease speed OR stop B1

change direction / move in a curve o.w.t.t.e

Explanation:

accelerate / increase speed OR decelerate / decrease speed OR stop B1

change direction / move in a curve o.w.t.t.e

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2 years ago

What are the steps of mitosis

Mice21 [21]

Prophase metaphase anaphase telophase

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2 years ago

You throw a glove straight upward to celebrate a victory. Its initial kinetic energy is K and it reaches a maximum height h. Wha

Rasek [7]

Answer:

K/2

Explanation:

The law of conservation of mechanical energy states that the sum of the kinetic and potential energies is a constant at any point.

At maximum height, the glove has purely potential energy but at the bottom, it has purely kinetic energy.

The potential energy at the top = kinetic energy at the bottom. The potential energy is given by

$PE = mgh$

At half height, this potential energy is

$PE = \frac{1}{2}mgh$

At this height, PE + KE = Constant = KE at bottom or PE at maximum height.

$mgh = \frac{1}{2}mgh +KE$

$KE = \frac{1}{2}mgh = K/2$

5 0

2 years ago

g A bowling ball with a mass of 3.86 kg and a radius of 0.161 m starts from rest at a height of 2.5 m and rolls down a 48.4 o sl

Fynjy0 [20]

Answer:

$v=1.5m/s$

Explanation:

The gravitational potential energy gets transformed into translational and rotational kinetic energy, so we can write $mgh=\frac{mv^2}{2}+\frac{I\omega^2}{2}$ . Since $v=r\omega$ (the ball rolls without slipping) and for a solid sphere $I=\frac{2mr^2}{5}$ , we have:

$mgh=\frac{mv^2}{2}+\frac{2mr^2\omega^2}{2*5}=\frac{mv^2}{2}+\frac{mv^2}{5}=\frac{7mv^2}{10}$

So our translational speed will be:

$v=\sqrt{\frac{10gh}{7}}=\sqrt{\frac{10(9.8m/s^2)(0.161m)}{7}}=1.5m/s$

6 0

2 years ago

Two scales on a nondigital voltmeter measure voltages up to 20.0 and 30.0 V, respectively. The resistance connected in series wi

Snowcat [4.5K]

Answer:

Resistance of the circuit is 820 Ω

Explanation:

Given:

Two galvanometer resistance are given along with its voltages.

Let the resistance is "R" and the values of voltages be 'V' and 'V1' along with 'G' and 'G1'.

⇒ $V=20\ \Omega,\ V_1=30\ \Omega$

⇒ $G=1680\ \Omega,\ G_1=2930\ \Omega$

Concept to be used:

Conversion of galvanometer into voltmeter.

Let $G$ be the resistance of the galvanometer and $I_g$ the maximum deflection in the galvanometer.

To measure maximum voltage resistance $R$ is connected in series .

So,

⇒ $V=I_g(R+G)$

We have to find the value of $R$ we know that in series circuit current are same.

For $G=1680$ For $G_1=2930$

⇒ $I_g=\frac{V}{R+G}$ equation (i) ⇒ $I_g=\frac{V_1}{R+G_1}$ equation (ii)

Equating both the above equations:

⇒ $\frac{V}{R+G} = \frac{V_1}{R+G_1}$

⇒ $V(R+ G_1) = V_1 (R+G)$

⇒ $VR+VG_1 = V_1R+V_1G$

⇒ $VR-V_1R = V_1G-VG_1$

⇒ $R(V-V_1) = V_1G-VG_1$

⇒ $R =\frac{V_1G-VG_1}{(V-V_1)}$

⇒ Plugging the values.

⇒ $R =\frac{(30\times 1680) - (20\times 2930)}{(20-30)}$

⇒ $R =\frac{(50400 - 58600)}{(-10)}$

⇒ $R=\frac{-8200}{-10}$

⇒ $R=820\ \Omega$

The coil resistance of the circuit is 820 Ω .

4 0

2 years ago