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wolverine [178]

4 years ago

6

Difinition of gravity

2 answers:

erastovalidia [21]4 years ago

6 0

Answer:

it's all around you and it can't be destroyed

kiruha [24]4 years ago

3 0

Answer: The force that attracts a body toward the center of the earth, or toward any other physical body having mass.

Explanation:

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A balloon rubbed against denim gains a charge of −7.0 µc. what is the electric force between the balloon and the denim when the

zubka84 [21]

The solution for this problem is computed by through this formula, F = kQq / d²Plugging in the given values above, we can now compute for the answer.
F = 8.98755e9N·m²/C² * -(7e-6C)² / (0.03m)² = -489N, the negative sign denotes attraction.

4 0

3 years ago

A .183 kg ball is moving 18.8 m/s when it runs into a spring of spring constant 86.9 N/m. How much KE does the ball have when it

Lemur [1.5K]

Answer:

The value is $KE_B = 20.59 \ J$

Explanation:

From the question we are told that

The mass of the ball is $m = 183 \ kg$

The initial speed of the ball is $u = 18.8 \ m/s$

The spring constant is $k = 86.9 \ N/m$

The compression distance is $x = 0.520 \ m$

Generally the energy stored in the string is mathematically represented as

$E = \frac{1}{2} * k * x^2$

=> $E = \frac{1}{2} * 86.9 * 0.520 ^2$

=> $E = 11.75 \ J$

Generally the kinetic energy of the ball is mathematically represented as

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

=> $KE_b = \frac{1}{2} * 0.183 * (18.8 )^2$

$KE_b = 32.34 \ J$

Generally the KE the ball have when it has compressed the spring is mathematically represented as

$KE_B = KE_b - E$

=> $KE_B = 32.34 - 11.75$

=> $KE_B = 20.59 \ J$

7 0

3 years ago

A wheel has a rotational inertia of 16 kgm2. Over an interval of 2.0 s its angular velocity increases from 7.0 rad/s to 9.0 rad/

german

Answer:

<h2>128.61 Watts</h2>

Explanation:

Average power done by the torque is expressed as the ratio of the workdone by the toque to time.

Power = Workdone by torque/time

Workdone by the torque = $\tau \theta$ = $I\alpha * \theta$

I is the rotational inertia = 16kgm²

$\theta = angular\ displacement$

$\theta = 2 rev = 12.56 rad$

$\alpha \ is \ the\ angular\ acceleration$

To get the angular acceleration, we will use the formula;

$\alpha = \frac{\omega_f^2- \omega_i^2}{2\theta}$

$\alpha = \frac{9.0^2- 7.0^2}{2(12.54)}\\\alpha = 1.28\ rad/s^{2}$

Workdone by the torque = 16 * 1.28 * 12.56

Workdone by the torque = 257.23 Joules

Average power done by the torque = Workdone by torque/time

= 257.23/2.0

= 128.61 Watts

8 0

3 years ago

What two properties show that a drink is a fluid

Kryger [21]

Two properties that show that the drink is a fluid/liquid include its retention of volume and its conformation to the shape of its container.

7 0

3 years ago

Liquid water at 25°C and 1 bar fills a rigid vessel. If heat is added to the water until its temperature reaches 50°C, what pres

Tcecarenko [31]

Answer:

The developed pressure is 205.75 bar.

Explanation:

Given that,

Temperature of water =25°C = 273+25 = 298 K

Pressure = 1 bar

Increases temperature = 50°C = 273+50 = 323 K

Value of $\beta = 36.2\times10^{-5}\ K^{-1}$

Value of $k=4.42\times10^{-5}\ bar^{-1}$

Specific volume of liquid = 1.0030 cm³/g

We need to calculate the final pressure

Using formula for constant volume change

$\beta(T_{2}-T)_{1})-k(P_{2}-P_{1})=0$

Put the value into the formula

$36.2\times10^{-5}(323-298)-4.42\times10^{-5}(P_{2}-1)=0$

$P_{2}=\dfrac{36.2\times10^{-5}(323-298)}{4.42\times10^{-5}}+1$

$P_{2}=205.75\ bar$

Hence, The developed pressure is 205.75 bar.

5 0

3 years ago