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lord [1]
2 years ago
7

A b c or d for not heat

Physics
2 answers:
Oksana_A [137]2 years ago
8 0
The awnser would be a
Levart [38]2 years ago
7 0

Answer:

a

Explanation:

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Calculate the kinetic energy of a 0.032 kg ball as it leaves a hand to be thrown upwards at 6.2 m/s
AnnZ [28]

Answer:

The ball will have a kinetic energy of 0.615 Joules.

Explanation:

Use the kinetic energy formula

E_k = \frac{1}{2}mv^2 = \frac{1}{2}0.032kg\cdot 6.2^2 \frac{m^2}{s^2}= 0.615J

The kinetic energy at the moment of leaving the hand will be 0.615 Joules. (From there on, as it ball is traveling upwards, this energy will be gradually traded off with potential energy until the ball's velocity becomes zero at the apex of the flight)

3 0
3 years ago
The roller coaster is designed according to safety regulations that prohibit the speed of the car from exceeding 20 m/s. find th
hram777 [196]

maximum allowed value of the speed in roller coaster is given as

v = 20 m/s

now from kinematics we can say

v^2 - v_i^2 = 2 a s

here initial speed will be

v_i = 0

acceleration is due to gravity

a = 9.8 m/s^2

now we can use this to find the height

20^2 - 0^2 = 2 * 9.8* h

400 = 19.6 *h

h = 20.4 m

so maximum allowed height will be 20.4 m

4 0
3 years ago
What does the ideal gas law allow a scientist to calculate that the other laws do not?
Daniel [21]
The ideal gas law.
PV=nRT
P=presure
V=volume
n=number of moles
R=Gas costant
T=temperature.

Answer: a. Number of moles.
6 0
2 years ago
Read 2 more answers
A 3.5 kg object moving in two dimensions initially has a velocity v1 = (12.0 i^ + 22.0 j^) m/s. A net force F then acts on the o
lys-0071 [83]

Answer:

The work done by the force is 820.745 joules.

Explanation:

Let suppose that changes in potential energy can be neglected. According to the Work-Energy Theorem, an external conservative force generates a change in the state of motion of the object, that is a change in kinetic energy. This phenomenon is describe by the following mathematical model:

K_{1} + W_{F} = K_{2}

Where:

W_{F} - Work done by the external force, measured in joules.

K_{1}, K_{2} - Translational potential energy, measured in joules.

The work done by the external force is now cleared within:

W_{F} = K_{2} - K_{1}

After using the definition of translational kinetic energy, the previous expression is now expanded as a function of mass and initial and final speeds of the object:

W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2})

Where:

m - Mass of the object, measured in kilograms.

v_{1}, v_{2} - Initial and final speeds of the object, measured in meters per second.

Now, each speed is the magnitude of respective velocity vector:

Initial velocity

v_{1} = \sqrt{v_{1,x}^{2}+v_{1,y}^{2}}

v_{1} = \sqrt{\left(12\,\frac{m}{s} \right)^{2}+\left(22\,\frac{m}{s} \right)^{2}}

v_{1} \approx 25.060\,\frac{m}{s}

Final velocity

v_{2} = \sqrt{v_{2,x}^{2}+v_{2,y}^{2}}

v_{2} = \sqrt{\left(16\,\frac{m}{s} \right)^{2}+\left(29\,\frac{m}{s} \right)^{2}}

v_{2} \approx 33.121\,\frac{m}{s}

Finally, if m = 3.5\,kg, v_{1} \approx 25.060\,\frac{m}{s} and v_{2} \approx 33.121\,\frac{m}{s}, then the work done by the force is:

W_{F} = \frac{1}{2}\cdot (3.5\,kg)\cdot \left[\left(33.121\,\frac{m}{s} \right)^{2}-\left(25.060\,\frac{m}{s} \right)^{2}\right]

W_{F} = 820.745\,J

The work done by the force is 820.745 joules.

6 0
2 years ago
What is the magnitude b of the magnetic field at the location of the charge due to the current-carrying wire?
elena-14-01-66 [18.8K]

Answer:

k

Explanation:

3 0
3 years ago
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