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Irina18 [472]
3 years ago
12

If you wanted to know the location of a vehicle that ran out of gas after taking a zigzag route through the city, which quantity

would be
most useful?

a. the vehicle's speed

b. the vehicle's distance

c. the vehicle's direction

d. the vehicle's displacement
Physics
1 answer:
Papessa [141]3 years ago
8 0

Explanation:

the vehicles displacement, since displacement deals with position

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If you fired a rifle straight upwards at 1000 m/s, how far up will the bullet get?
nadya68 [22]

Answer:

h = 51020.40 meters

Explanation:

Speed of the rifle, v = 1000 m/s

Let h is the height gained by the bullet. It can be calculated using the conservation of energy as :

\dfrac{1}{2}mv^2=mgh

h=\dfrac{v^2}{2g}

h=\dfrac{(1000\ m/s)^2}{2\times 9.8\ m/s^2}                    

h = 51020.40 meters

So, the bullet will get up to a height of 51020.40 meters. Hence, this is the required solution.          

4 0
2 years ago
Based on the data in the table...especially the provided melting points, which two substances are
ArbitrLikvidat [17]

The two substances that are mostly likely examples of covalent bonding are Sucrose and Ethanol.

<h3 /><h3 /><h3>What is a covalent Bond?</h3>
  • A covalent bond is a type of chemical bond that involves the sharing of pairs of electron between atoms.

Examples of compounds with covalent bond include the following;

  • Distilled water
  • Sucrose
  • Ethanol

Olive oil is a mixture not a compound

Sodium Chloride & Potassium lodide are examples of ionic bond.

Thus, the two substances that are mostly likely examples of covalent bonding are Sucrose and Ethanol.

Learn more about covalent bonds here: brainly.com/question/12732708

7 0
2 years ago
How much heat is required to heat 2 kg of water from 25°C to 40°C?
Dominik [7]

Answer:

126000 J

Explanation:

Applying,

Q = cm(t₂-t₁).................. Equation 1

Where Q = Amount of heat, c = specifc heat capacity of water, m = mass of water, t₁ = Initial temperature, t₂ = Final temperature.

From the question,

Given: m = 2 kg, t₁ = 25°C, t₂ = 40°C

Constant: c = 4200 J/kg.°C

Substitute these value into equation 1

Q = 2×4200(40-25)

Q = 2×4200×15

Q = 126000 J

5 0
3 years ago
Suppose the ring rotates once every 4.30 ss . If a rider's mass is 58.0 kgkg , with how much force does the ring push on her at
Stells [14]

Answer:

422.36 N

Explanation:

given,

time of rotation = 4.30 s

T = 4.30 s

Assuming the diameter of the ring equal to 16 m

radius, R = 8 m

v = \dfrac{2\pi R}{T}

v = \dfrac{2\pi\times 8}{4.30}

  v = 11.69 m/s

now, Force does the ring push on her at the top

- N - m g = \dfrac{-mv^2}{R}

N + m g = \dfrac{mv^2}{R}

N = \dfrac{mv^2}{R}- m g

N = m(\dfrac{v^2}{R}- g)

N = 58\times (\dfrac{11.69^2}{8}- 9.8)

N = 422.36 N

The force exerted by the ring to push her is equal to 422.36 N.

6 0
2 years ago
If you throw a ball down ward then acceleration immeditely after leaving your hand is
bezimeni [28]

Answer:

9.8m/s²

Explanation:

The acceleration of the ball thrown after leaving my hand is 9.8m/s². This will be the acceleration due to gravity on the body.

  • Acceleration due to gravity is caused by the pull of the earth on a massive object.
  • The value of this acceleration is 9.8m/s².
  • As the ball nears the surface, it comes near zero.
7 0
2 years ago
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