Answer:
Collect more fuel than you think you can use; you may need more than you estimate. Pile fine twigs, grass, or bark shavings loosely as a base.
When the ball is first thrown, its kinetic energy is highest and its potential energy is lowest. As the ball rises, its kinetic energy decreases and its potential energy increases because it's slowing down as it goes higher. As the ball goes as high as it can, it stops momentarily; that is when the ball's potential energy is full, and it's kinetic energy is low. Then when the ball begins to fall back down, its potential energy is getting lower, and its kinetic energy is growing. Right before the ball hit the ground, the kinetic energy was at its highest, and its potential was lowest. Then when it hit the ground, there was no kinetic or potential energy.
12.) Active transport because the cell must use energy to move large particles across the membrane.
13.) Photosynthesis takes place in plant leaves containing the chlorophyll pigment. Cellular respiration takes place in the cytoplasm and mitochondria of the cell. ... Cellular respiration uses glucose molecules and oxygen to produce ATP molecules and carbon dioxide as the by-product.
14.) In cells with a nucleus, as in eukaryotes, the cell cycle is also divided into two main stages: interphase and the mitotic (M) phase (including mitosis and cytokinesis). During interphase, the cell grows, accumulating nutrients needed for mitosis, and undergoes DNA replication preparing it for cell division.
323.5 N is the tension in the cable.
Given
Mass of crate(M) = 175.5 kg
Mass of boom(m) = 94.7 kg
The tension, T in the cable can be calculated by taking moments of force about the central point of marked X.
The Angle of the boom with the horizontal can be calculated by
tanθ = 5/10
θ = tan⁻¹(5/10) = 26.56°
Angle of the boom with horizontal is 26.56°
The angle of cable with horizontal can be calculated by
tan B = 4/10
B = tan⁻¹(4/10) = 21.80°
Angle of cable with horizontal is 21.80°
Taking moments of force about the point X
(Mcosθ + mcosθ) 0.5 = T(sin(θ +B)1
(175.5 × cos 26.56 + 94.7 × cos 26.56 )× 0.5 = T (sin(26.56 + 21.80) X 1
By calculating, we get
Tension(T) = 241.68/0.747
Tension(T) = 323.5 N
Hence, 323.5 N is the tension in the cable.
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