B. inversely proportional
C. have no effect on each other
In an electric circuit, resistance and current are ____
A. directly proportional
B. inversely proportional
C. have no effect on each other
Explanation:
A
A current of 67 amps runs through a resistor of 37 ohms, how much voltage is lost?
Dawn is trying to figure out how much weight she can push with her strength, or what her maximum pushing force is, across the room. She could do an experiment to find out.
She must first prepare a testing space with a flat, smooth surface to reduce friction. She can then progressively add weights to a cart or other object and use all of her strength to try to push it across the room. She can determine her maximum pushing force by noting the heaviest weight she can move. For a variety of jobs, including moving furniture or participating in physical sports that call for pushing heavy things, this knowledge can be essential.
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Answer:
Muscular strength
Explanation:
She is testing her strength while pushing the weights
Answer:
A. 29.7 m/s
B. 6.06 s
Explanation:
From the question given above, the following data were obtained:
Maximum height (h) = 45 m
A. Determination of the initial velocity (u)
Maximum height (h) = 45 m
Acceleration due to gravity (g) = 9.8 m/s²
Final velocity (v) = 0 m/s (at maximum height)
Initial velocity (u) =.?
v² = u² – 2gh (since the ball is going against gravity)
0² = u² – (2 × 9.8 × 45)
0 = u² – 882
Collect like terms
0 + 882 = u²
882 = u²
Take the square root of both side
u = √882
u = 29.7 m/s
Therefore, the ball must be thrown with a speed of 29.7 m/s.
B. Determination of the time spent by the ball in the air.
We'll begin by calculating the time take to reach the maximum height. This can be obtained as follow:
Maximum height (h) = 45 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) to reach the maximum height =?
h = ½gt²
45 = ½ × 9. 8 × t²
45 = 4.9 × t²
Divide both side by 4.9
t² = 45/4.9
Take the square root of both side
t = √(45/4.9)
t = 3.03 s
Finally, we shall determine the time spent by the ball in the air. This can be obtained as follow:
Time (t) to reach the maximum height = 3.03 s
Time (T) spent by the ball in the air =?
T = 2t
T = 2 × 3.03
T = 6.06 s
Therefore, the ball spent 6.06 s in the air.
B. 60 cm
C. 75 cm
D. 90 cm
(a) It takes approximately 2.8956 seconds for the camera to reach the ground.
(b) The velocity of the camera just before it lands is approximately -28.375 m/s (downward).
We have,
Given information:
Initialheight (h₀) = 41 m (above the ground)
Descentrate of the hot-air balloon = -2.3 m/s (negative because it's descending)
We can use the kinematicequations to solve for the time it takes for the camera to reach the ground and its velocity just before landing.
(a)
To find the time it takes for the camera to reach the ground, we can use the following kinematic equation:
h = h₀ + (v₀)t + (1/2)at²
Where:
h = final height (0 m, as it reaches the ground)
h₀ = initial height (41 m)
v₀ = initial velocity (0 m/s, as the camera is dropped)
a = acceleration (acceleration due to gravity, approximately -9.8 m/s²)
t = time (what we're solving for)
Plugging in the values:
0 = 41 + (0)t + (1/2)(-9.8)t²
Simplifying:
-4.9t² = 41
Divide by -4.9:
t² = -41 / -4.9
t² = 8.36734694
Taking the square root:
t = √8.36734694
t ≈ 2.8956 seconds
(b)
To find the velocity just before the camera lands, we can use the following kinematic equation:
v = v₀ + at
Where:
v = final velocity (what we're solving for)
v₀ = initial velocity (0 m/s)
a = acceleration (acceleration due to gravity, -9.8 m/s²)
t = time (2.8956 seconds, calculated in part a)
Plugging in the values:
v = 0 + (-9.8)(2.8956)
v ≈ -28.375 m/s
The negativesign indicates that the velocity is directed downward, which is consistent with the descending motion.
Thus,
(a) It takes approximately 2.8956 seconds for the camera to reach the ground.
(b) The velocity of the camera just before it lands is approximately -28.375 m/s (downward).
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