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2 years ago

Which structures protect the cell? Check all that apply.

Physics

2 answers:

MAXImum [283]2 years ago

8 0

Answer:

cell wall and cell membrane

zimovet [89]2 years ago

6 0

Answer:

Cell wall and Cell membrane.

Explanation:

The cell membrane controls what goes in and out of the cell

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In a lab, the mass of object a is 52 kg. object a weighs

aksik [14]

If you are asking for the weight then the formula is F=mg where f is weight m is mass and g is acceleration due to gravity.m=52kg and g=9.8m/s2(the gravity of earth)
F=52*9.8=509.6
therefore the weight of the object is 509.6N

5 0

2 years ago

Read 2 more answers

In a high school swim competition, a student takes 1.6 s to complete 1.5 somersaults. Determine the average angular speed of the

Rufina [12.5K]

Answer:

The angular speed is $w = 5.89 \ rad/s$

Explanation:

From the question we are told that

The time taken is $t = 1.6 s$

The number of somersaults is n = 1.5

The total angular displacement during the somersault is mathematically represented as

$\theta = n * 2 * \pi$

substituting values

$\theta = 1.5 * 2 * 3.142$

$\theta = 9.426 \ rad$

The angular speed is mathematically represented as

$w = \frac{\theta }{t}$

substituting values

$w = \frac{9.426}{1.6}$

$w = 5.89 \ rad/s$

3 0

2 years ago

Sound is an example of a ______ wave.

tester [92]

Sound is a longitudinal wave.

7 0

2 years ago

Read 2 more answers

Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit

Ber [7]

Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

$= 822486 \times 10^{-8}$

$=0.822 \times 10^{-2} \ km/s$

= 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

$= 673552 \times 10^{-8}$

$=0.673 \times 10^{-2} \ km/s$

= 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

$= 3371 \times 153.86 \times 10^{-8}$

= 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

$= 153.84 \times 2371 \times 10^{-8}$

$=0.364 \times 10^{-2} \ km/s$

= 3.64 m/s

3 0

2 years ago

Opal adds 25 grams of salt to a one-liter glass beaker filled up to its volume mark with pure water. She stirs the water until t

Mars2501 [29]

By what i know i think that the answer would be A a homogeneous mixture.

5 0

2 years ago