Answer:
The student with the 3ft charger
Explanation:
Because the electricity from the 3ft charger has a shorter distance to mooves than the 9ft one
Answer:
Explanation:
Cylinder P is taller than cylinder Q. In fact, it's twice as tall. So if cylinder Q is, let's say, 10 inches tall, cylinder P would be 20 inches tall.
Cylinder P is also wider, but not by a lot. It's only half as wide as cylinder Q. So if cylinder Q has a width of 10 inches, cylinder P would have a width of 5 inches.
Now, let's talk about the insides of these cylinders, how much stuff they can hold. If we filled them up with something, like water, the bigger one (cylinder P) can hold twice as much as the smaller one (cylinder Q).
So, cylinder P can hold more stuff because it's both taller and a little wider than cylinder Q.
B. Green
C. Yellow
D. Red
Explain your reasoning.
With what minimum horizontal velocity must the cliff divers leave the top of the cliff if they are to miss the rock?
Answer in units of $m / s$.
Answer:
To solve this problem, we can use the kinematic equation for horizontal motion, which relates the initial velocity ($v_{0}$), final velocity ($v_{f}$), acceleration ($a$), and displacement ($d$) of an object:
$d = v_{0} t + \frac{1}{2}at^{2}$
In this case, we want to find the minimum initial velocity ($v_{0}$) that the divers must have to clear the rock. To do this, we can assume that the divers just graze the rock at the start of their trajectory, so the displacement in the horizontal direction is equal to the distance from the cliff to the rock ($d = 9.34 m$). We also know that the acceleration in the horizontal direction is zero, so the only force acting on the divers is gravity in the vertical direction, which gives an acceleration of $a = 9.8 m/s^{2}$.
At the instant the divers leave the cliff, they have zero horizontal velocity, so $v_{0} = 0$. We can use the equation above to solve for the time it takes for the divers to fall from the cliff to the level of the rock:
$d = \frac{1}{2}at^{2} \Rightarrow t = \sqrt{\frac{2d}{a}}$
Plugging in the numbers, we get:
$t = \sqrt{\frac{2(9.34 m)}{9.8 m/s^{2}}} \approx 1.44 s$
Since the cliff divers want to clear the rock, they need to travel a horizontal distance of at least $9.34 m$ during this time. We can use the equation for horizontal motion again to solve for the minimum initial velocity:
$d = v_{0}t \Rightarrow v_{0} = \frac{d}{t} = \frac{9.34 m}{1.44 s} \approx 6.49 m/s$
Therefore, the minimum horizontal velocity that the cliff divers must have to clear the rock is approximately $6.49 m/s$.
Answer:
Explanation:
To find the momentum of the recoiling particle you can use the momentum formula for a photon:
before the decay the momentum is zero. Hence, after the decay the momentum of the photon plus the momentum of the recoiling particle must be zero:
where pr is the momentum of the recoiling particle.
B.6.25 times heavier than on Earth
C.2.5 times heavier than on Earth
D.6.25 times lighter than on Earth
Answer: The correct answer is option C.
Explanation:
Weight = Mass × Acceleration
Let the mass of the space probe be m
Acceleration due to gravity on the earth = g
Weight of the space probe on earth = W
Acceleration due to gravity on the Jupiter = g' = 2.5g
Weight of the space probe on earth = W'
The weight of the space probe on the Jupiter will be 2.5 times the weight of the space probe on earth.
Hence, the correct answer is option C.
Answer:
2.5 times heavier than on Earth
Explanation: