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Alexxandr [17]
3 years ago
7

Hunter wants to measure the weight of a dumbbell. He should

Physics
2 answers:
MissTica3 years ago
4 0
Use a scale and record the weight in cm^3
Georgia [21]3 years ago
4 0

Answer:

use a scale and record the weight in N.

Explanatiuse a scale and record the weight in N.on:

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Assume the ground is uniformly level. If the horizontal component a projectile's velocity is doubled, but the vertical component
Katarina [22]

Answer:

Explanation:

Time of flight = 2 x u sinα / g where u sinα is vertical component of projectile's velocity u .

So Time of flight = 2 x vertical component / g

vertical component = constant

g is also constant so

Time of flight will also be constant .

It will remain unchanged .

3 0
3 years ago
What is the gravitational field theory and how does it work? four mark answer. Don't give me random answers just for points or i
MrMuchimi
It is a theory on a show that people try to solve.
6 0
2 years ago
Ball A 1.55kg moving right at 8.76 m/s makes a head-on collision with ball B (0.752 kg) moving left at 11.4 m/s. After, ball B m
Illusion [34]

1.15 m/s to the left (3 sig. fig.).

<h3>Explanation</h3>

Momentum is conserved between the two balls if they are not in contact with any other object. In other words,

p_{\text{A,initial}} + p_{\text{B,initial}}=p_{\text{A,final}} + p_{\text{B,final}}

m_\text{A} \cdot v_{\text{A,initial}} + m_\text{B}\cdot v_{\text{B,initial}}=m_\text{A}\cdot v_{\text{A,final}} + m_\text{B}\cdot v_{\text{B,final}}, where

  • m stands for mass and
  • v stands for velocity, which can take negative values.

Let the velocity of objects moving to the right be positive.

  • m_\text{A} = 1.55\;\text{kg},
  • m_\text{B} = 0.752\;\text{kg}.

Before the two balls collide:

  • v_\text{A} = +8.76\;\text{m}\cdot\text{s}^{-1},
  • v_\text{B} = -11.4\;\text{m}\cdot\text{s}^{-1}.

After the two balls collide:

  • v_\text{A} needs to be found,
  • v_\text{B} = +9.03\;\text{m}\cdot\text{s}^{-1}.

Again,

m_\text{A} \cdot v_{\text{A,initial}} + m_\text{B}\cdot v_{\text{B,initial}}=m_\text{A}\cdot v_{\text{A,final}} + m_\text{B}\cdot v_{\text{B,final}},

1.55 \times (+8.76) + 0.752 \times (-11.4) = 1.55\;{\bf v_{\textbf{A,final}}} + 0.752 \times (+9.03).

v_{\text{A,final}} = \dfrac{1.55 \times (+8.76) + 0.752 \times (-11.4)-0.752 \times (+9.03)}{1.55} = -1.15\;\text{m}\cdot\text{s}^{-1}.

v_{\text{A,final}} is negative? Don't panic. Recall that velocities to the right is considered positive. Accordingly, negative velocities are directed to the left.

Hence, ball A will be travelling to the left at 1.15 m/s (3 sig. fig. as in the question) after the collision.

8 0
3 years ago
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An object travels with velocity v = 4.0 meters/second and it makes an angle of 60.0° with the positive direction of the y-axis. 
Darya [45]
Goooooooaaaaalllll , C 2.5 meters
7 0
3 years ago
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If the temperature at the surface of Earth (at sea level) is 100°F, what is the temperature at 2000 feet if the average lapse ra
oksano4ka [1.4K]

Answer:

107 °F

Explanation:

Given that

The temperature at sea level = 100°F

height ,h= 2000 feet

The average lapse rate = 3.5°F/1000 feet

Given that rise in temperature 3.5°F per 1000 feet.

1000 feet ⇒ 3.5°F

Given that 2000 feet

2000 feet ⇒ 3.5°F x 2 +100°F

2000 feet ⇒ 107 °F

Therefore the temperature will be 107 °F  .

6 0
3 years ago
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