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Paha777 [63]
2 years ago
8

A dart is thrown horizontally at a target's center that is 5.00\,\text m5.00m5, point, 00, start text, m, end text away. The dar

t hits the target 0.150\,\text m0.150m0, point, 150, start text, m, end text below the target's center.What was the initial horizontal velocity of the dart?
Physics
2 answers:
Sholpan [36]2 years ago
8 0

Answer:

Explanation:

Given

the range = 5.00m (distance moved in horizontal direction)

Height = 0.150m

Required

Initial velocity of the dart

Using the formula for calculating range of a projectile;

R = U√2H/g

5 = U√2(0.15)/9.8

5 = U√0.0306

5 = 0.1749U

U = 5/0.1749

U = 28.59m/s

Hence the initial horizontal velocity of the dart is 28.59m/s

alexandr402 [8]2 years ago
6 0

Answer:

28.59

Explanation:

khan academy

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Answer:

Friction:  is used to hang an object on the wall

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Explanation:

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2 years ago
I WILL MARK BRAINLIEST!!ASAP!!! Wet Lab - Coulomb's Law lab from edge!!
snow_tiger [21]

Answer:

h

Explanation:

Coulomb's law, or Coulomb's inverse-square law, is an experimental law[1] of physics that quantifies the amount of force between two stationary, electrically charged particles. The electric force between charged bodies at rest is conventionally called electrostatic force or Coulomb force.[2] The law was first discovered in 1785 by French physicist Charles-Augustin de Coulomb, hence the name. Coulomb's law was essential to the development of the theory of electromagnetism, maybe even its starting point,[1] as it made it possible to discuss the quantity of electric charge in a meaningful way.[3]

The law states that the magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them,[4]

{\displaystyle F=k_{\text{e}}{\frac {q_{1}q_{2}}{r^{2}}}}{\displaystyle F=k_{\text{e}}{\frac {q_{1}q_{2}}{r^{2}}}}

Here, ke is Coulomb's constant (ke ≈ 8.988×109 N⋅m2⋅C−2),[1] q1 and q2 are the signed magnitudes of the charges, and the scalar r is the distance between the charges.

The force is along the straight line joining the two charges. If the charges have the same sign, the electrostatic force between them is repulsive; if they have different signs, the force between them is attractive.

Being an inverse-square law, the law is analogous to Isaac Newton's inverse-square law of universal gravitation, but gravitational forces are always attractive, while electrostatic forces can be attractive or repulsive.[2] Coulomb's law can be used to derive Gauss's law, and vice versa. In the case of a single stationary point charge, the two laws are equivalent, expressing the same physical law in different ways.[5] The law has been tested extensively, and observations have upheld the law on the scale from 10−16 m to 108 m.[5]

7 0
3 years ago
A baseball with a mass of 300 g has a kinetic energy of 304 J. Calculate the speed of the baseball
scZoUnD [109]

The formula for kinetic energy is equal to 1/2mv^2, where "m" is the mass of the object (in kilograms) and "v" is equal to the velocity of the object (in meters per second). To calculate the speed, simply plug in the values and solve.

KE = 0.5mv^2

304 J = 0.5(0.3 kg)v^2         -mass converted from grams to kilograms

v = 45.02 m/s

The baseball is travelling about 45.02 meters per second.

Hope this helps!

4 0
3 years ago
Read 2 more answers
The pressure in a compressed air storage tank is 1200 kPa. What is the tank’s pressure in (a) kN and m units; (b) kg, m, and s u
elixir [45]

Explanation:

Given data

Pressure P=1200 kPa

To find

Pressure in

(a) kN/m²

(b) kg/m.s²

(c) kg/km.s²

Solution

For Part (a)

P=(1200kPa)(\frac{1kN/m^{2} }{1kPa} )\\P=1200kN/m^{2}

For Part (b)

P=(1200kPa)(\frac{1kN/m^{2} }{1kPa} )(\frac{1000kg.m/s^{2} }{1kN} )\\P=1,200,000kg/m.s^{2}

For Part (c)

P=(1200kPa)(\frac{1kN/m^{2} }{1kPa} )(\frac{1000kg.m/s^{2} }{1kN} )(\frac{100m}{1km} )\\P=1,200,000,000kg/km.s^{2}

8 0
3 years ago
When two point charges are a distance d part, the electric force that each one feels from the other has magnitude F. In order to
Contact [7]

Answer:

Because the force is inversely proportional to the square of the distance

Explanation:

The magnitude of the electrostatic force between two charged particles is given by

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

where

k is the Coulomb's constant

q1, q2 are the magnitudes of the two charges

d is the distance between the two charges

We observe that the magnitude of the force is inversely proportional to the square of the distance.

Therefore, when the distance changes to

d'=\frac{d}{\sqrt{2}}

The force will double:

F'=k\frac{q_1 q_2}{(d/\sqrt{2})^2}=2(k\frac{q_1 q_2}{d^2})=2F

5 0
3 years ago
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