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3 years ago

Identifying Structures in Photosynthesis

Physics

2 answers:

n200080 [17]3 years ago

8 0

Answer:

Explanation:

OLEGan [10]3 years ago

4 0

I believe the answer is 1 sorry if I’m wrong

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In a downhill ski race, surprisingly, little advantage is gained by getting a running start. (This is because the initial kineti

Sergio [31]

Answer:

Part a)

$v_f = 25.2 m/s$

$t = 5.48 s$

Part b)

$v_f = 25.32 m/s$

$t = 4.96 s$

Explanation:

Part a)

When ski start from rest

$v_f^2 - v_i^2 = 2 a d$

on this inclined plane we know that the acceleration is given as

$a = g sin\theta$

$a = 9.81 sin28$

$a = 4.6 m/s^2$

now for final speed

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 0 = 2(4.6)(69)$

$v_f = 25.2 m/s$

now time taken by the ski to reach the bottom is given as

$v_f = v_i + at$

$25.2 = 0 + 4.6 t$

$t = 5.48 s$

Part b)

Now when ski start with initial speed of 2.5 m/s

then we will have

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 2.5^2 = 2(4.6)(69)$

$v_f = 25.32 m/s$

now time taken by the ski to reach the bottom is given as

$v_f = v_i + at$

$25.32 = 2.5 + 4.6 t$

$t = 4.96 s$

3 0

3 years ago

A runner starts at position A. He runs 40 m North, 10 m East and 40 m

Alecsey [184]

The answer is C.) 10 m East

8 0

3 years ago

A 320-g ball and a 400-g ball are attached to the two ends of a string that goes over a pulley with a radius of 8.7 cm. Because

Gnom [1K]

To solve this problem, it is necessary to apply the concepts related to force described in Newton's second law, so that

F = ma

Where,

m = mass

a = Acceleration (Gravitational acceleration when there is action over the object of the earth)

Torque, as we know, is the force applied at a certain distance, that is,

$\tau = F*d$

Where

F= Force

d = Distance

Our values are given as,

$m_1 = 0.32Kg$

$m_2 = 0.4Kg$

$d = 8.7*10^{-2}m$

Since the system is in equilibrium the difference of the torques is the result of the total Torque applied, that is to say

$\tau = T_2-T_1$

$\tau = F_2*d-F_1*d$

$\tau = m_2g*d-m_1*g*d$

$\tau = (m_2-m_1)g*d$

$\tau = (0.4-0.32)(9.8)(8.7*10^{-2})$

$\tau = 0.068N\cdot m$

Therefore the magnitude of the frictional torque at the axle of the pulley if the system remains at rest when the balls are released is $\tau = 0.068N\cdot m$

8 0

3 years ago

A frictionless piston–cylinder device contains 5 kg of nitrogen at 100 kPa and 250 K. Nitrogen is now compressed slowly accordin

Arte-miy333 [17]

Answer:

The work input during this process is -742 kJ

Explanation:

Given;

Initial temperature of nitrogen T₁ = 250 K

final temperature of nitrogen T₂ = 450 K

mass of nitrogen, m = 5 kg

$PV^{1.4} = constant$

The work input during the process is calculated as;

$W = \frac{m*R(T_2-T_1)}{1-n}$

where;

R is gas constant = 0.2968 kJ/kgK

substitute given values in above equation.

$W = \frac{m*R(T_2-T_1)}{1-n} = \frac{5*0.2968(450-250)}{1-1.4} = -742 \ kJ$

Therefore, the work input during this process is -742 kJ

8 0

4 years ago

What might be done to prevent acid rain damage to objects made from metal and carbonate, such as limestone? Identify a metal obj

serg [7]

Answer: Bronze is a metal that can be damaged in acid rain. It is harsh to metals by almost burning them and may create rust. Everbrite is a clear, protective coating that will seal metal and bronze.

Explanation:

I know this is correct i have done all the research

7 0

3 years ago