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lapo4ka [179]
3 years ago
11

i will give brainliest Which of the following structures are present in plant cells but absent in animal cells? Choose 1 answer:

(Choice A) A Cell walls (Choice B) B Mitochondria (Choice C) C Cytosol (Choice D) D Golgi bodies
Physics
1 answer:
Sonja [21]3 years ago
3 0

Answer:

A. cell walls

Explanation:

Plants have cell walls but animals dont.

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Neither the speed nor the distance of a falling object is linearly related to time.
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Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at c is 3.39 m/s and
djverab [1.8K]

The magnitude of the average friction force exerted on the collar (F)=  8.641 N

<h3>How can we calculate the magnitude of the average friction force exerted on the collar?</h3>

To calculate the magnitude of the average friction force exerted on the collar we are using the formula,

\frac{1}{2} k(x^2_f - x^2_i ) + F\times y + \frac{1}{2} m v^{2} _{c}  = mgy

Here we are given,

k = The spring has a spring constant.

= 25.5 N/m.

x_f = Final length of the spring .

= \sqrt{1.25^2+1.8^2}  -0.60

= 1.591 m

x_i= The initial length of the spring.

= 1.25−0.60

=0.65 m

y=The collar then travels downward a distance.

=  1.80 m.

m= The mass of the collar.

=3.55 kg

v_c = the velocity of the collar.

= 3.39 m/s.

g = The acceleration due to gravity.

= 9.81 m/s²

We have to calculate the magnitude of the average friction force exerted on the collar = F

Now we put the known values in the above equation, we get;

\frac{1}{2} k(x^2_f - x^2_i ) + F\times y + \frac{1}{2} m v^{2} _{c}  = mgy

Or, \frac{1}{2} \times 25.5 \times((1.591)^2 - (0.65)^2 ) + F\times 1.80 + \frac{1}{2}\times 3.55\times (3.39)^{2}  = 3.55\times 9.81\times 1.80

Or, F= 8.641 N

From the above calculation we can conclude that,

The magnitude of the average friction force exerted on the collar (F)=  8.641 N

Learn more about friction:

brainly.com/question/24338873

#SPJ4

Disclaimer: This question is incomplete in the portal. Here is the complete question.

Question:

The 3.55 kg collar shown below is attached to a spring and released from rest at A. The collar then travels downward a distance of y = 1.80 m. The spring has a spring constant of k = 25.5 N/m. The distance a is given as 1.25 m. The datum for gravitational potential energy is set at the horizontal line through A and B. Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at c is 3.39 m/s and the spring has an unstretched length of 0.60 m .

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Adding resistors in series changes the total resistance of a circuit by
skelet666 [1.2K]

Answer:

Increasing the resistance

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A ball is kicked at a speed of 16m/s at 33° and it eventually returns to ground level further down field.
saw5 [17]

Hi there!

We can begin by calculating the time taken to reach its highest point (when the vertical velocity = 0).

Remember to break the velocity into its vertical and horizontal components.

Thus:

0 = vi - at

0 = 16sin(33°) - 9.8(t)

9.8t = 16sin(33°)

t = .889 sec

Find the max height by plugging this time into the equation:

Δd = vit + 1/2at²

Δd = (16sin(33°))(.889) + 1/2(-9.8)(.889)²

Solve:

Δd = 7.747 - 3.873 = 3.8744 m

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Consider a sample of gas in a container on a comfortable spring day in chicago, il. the celsius temperature suddenly doubles, an
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To solve this problem, we must first assume that the gas acts like an ideal gas so that we can use the ideal gas equation:

 P V = n R T

where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant and T is the absolute temperature

 

Assuming that the number of moles is constant, then we can write all the variables in the left side:

P V / T = k            where k is a constant (n times R)

 

Equating two conditions or two states:

P1 V1 / T1 = P2 V2 / T2

We are given that V2 = 2 V1 therefore

P1 V1 T2 = P2 (2V1) T1

P1 T2 = 2 P2 T1

 

Additionally we are given that the temperature in Celsius is doubled, however in the formula we use the absolute temperature in Kelvin, therefore:

T1 (K) = T1 + 273.15

T2 (K) = 2T1 + 273.15

and P1 = 12 atm

 

Substituting:

<span>12 (2T1 + 273.15)  = 2 P2 (T1 + 273.15)</span>

P2 = 6 (2T1 + 273.15) / (T1 + 273.15)

 

Assuming that a nice spring day in Chicago has a temperature of 15 Celsius, therefore:

P2 = 6 (2*15 + 273.15) / (15 + 273.15)

<span>P2 = 6.312 atm</span>

3 0
3 years ago
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