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3 years ago

98 Points and brainlyest for 5 Science questions please I need it doe before 2:30 ET!!!

Physics

1 answer:

Marina CMI [18]3 years ago

5 0

Picture #1:
GPE = (mass) x (gravity) x (height)
GPE = (2 kg) x (9.8 m/s²) x (40 m) = 784 joules

KE = (1/2) (mass) (speed²)
KE = (1/2) (2 kg) (5 m/s)²
KE = (1 kg) (25 m²/s²) = 25 joules

Picture #2:
KE = (1/2) (mass) (speed²)
KE = (1/2) (2 kg) (10 m/s)²
KE = (1 kg) (100 m²/s²) = 100 joules

Picture #3:
GPE = (mass) x (gravity) x (height)
GPE = (20 kg) x (9.8 m/s²) x (2 m) = 392 joules

KE = (1/2) (mass) (speed²)
KE = (1/2) (20 kg) (5 m/s)²
KE = (10 kg) (25 m²/s²) = 250 joules

Picture #4:
GPE = (mass) x (gravity) x (height)
98 joules = (1 kg) x (9.8 m/s²) x (height)
Height = (98 joules) / (1 kg x 9.8 m/s²)
Height = 10 meters

Picture #5:
GPE = (mass) x (gravity) x (height)
39,200 Joules = (mass) x (9.8 m/s²) x (20 m)
Mass = (39,200 joules) / (9.8 m/s² x 20 m)
Mass = 200 kg

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Ipatiy [6.2K]

Explanation:

A force that leads to movement of an object is known as work.

The energy present in an object due to its position in a gravitational field is known as gravitational potential energy.

Kinetic energy is the energy obtained by an object due to its motion.

For example, when Jerome is swinging on a rope then there occurs movement in the swing due to which the swing has kinetic energy.

Since, a force has been applied on the swing to make it move. Hence, a work is also done.

Therefore, we can conclude that if Jerome is swinging on a rope and transferring energy from gravitational potential energy to kinetic energy, work is being done.

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3 years ago

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three point charges are arranged in a line. charge q3=+5.00 nC and is located at the origin. charge q2=-3.00 nC and is located a

valentinak56 [21]

Answer:

$q_1=+0.375\ {10}^{-9}$

Explanation:

Electrostatic Forces

The force exerted between two point charges $q_1$ and $q_2$ separated a distance d is given by Coulomb's formula

$\displaystyle F=\frac{k\ q_1\ q_2}{d^2}$

The forces are attractive if the charges have different signs and repulsive if they have equal signs.

The problem described in the question locates three point charges in a straight line. The charges have the values shown below

$\displaystyle q_3=+5\ 10^{-9}\ c$

$\displaystyle q_2=-3\ 10^{-9}\ c$

The distance between $q_3$ and $q_2$ is

$\displaystyle d_2=4cm=0.04\ m$

The distance between $q_3$ and $q_1$ is

$\displaystyle d_1=2cm=0.02\ m$

We must find the value of $q_1$ such that

$\displaystyle |F_3|=0$

Applying Coulomb's formula for $q_1$ is

$\displaystyle F_{13}=\frac{k\ q_1\ q_3}{d_1^2}$

Now for $q_2$

$\displaystyle F_{23}=\frac{k\ q_2\ q_3}{d_2^2}$

If the total force on $q_3$ is zero, both forces must be equal. Note that being q2 negative, the force on q3 is to the right. The force exerted by q1 must go to the left, thus q1 must be positive. Equating the forces we have:

$\displaystyle F_{13}=F_{23}$

$\displaystyle \frac{k\ q_1\ q_3}{d_1^2}=\frac{k\ q_2\ q_3}{d_2^2}$

Simplfying and solving for $q_1$

$\displaystyle q_1=\frac{q_2\ d_1^2}{d_2^2}$

$\displaystyle q_1=\frac{3.10^{-9}\ 0.02^2}{0.04^2}$

$\boxed{\displaystyle q_1=+0.375\ {10}^{-9}}$

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3 years ago

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Answer:

The total work done by the two tugboats on the supertanker is 3.44 *10^9 J

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The force by the tugboats acting on the supertanker is constant and the displacement of the supertanker is along a straight line.

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First we have to calculate the work done by the individual force and then we can calculate the total work.

The work done on a particle by a constant force F during a straight line displacement s is given by following formula:

W = F*s

W = F*s*cos∅

With ∅ = the angles between F and s

The magnitude of the force acting on the supertanker is F of tugboat1 = F of tugboat 2 = F = 2.2 * 10^6 N

The total work done can be calculated as followed:

Wtotal = Ftugboat1 s * cos ∅1 + Ftugboat2 s* cos ∅2

Wtotal = 2Fs*cos∅

Wtotal = 2*2.2*10^6 N * 0.81 *10³ m s *cos15°

Wtotal = 3.44*10^9 Nm = 3.44 *10^9 J

The total work done by the two tugboats on the supertanker is 3.44 *10^9 J

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