Which cellular structure would be the most likely location for the synthesis of these enzymes?

babymother [125]

Answer:

11,100 N

Explanation:

Draw the car on the incline. Label the weight and normal forces. Call the direction normal to the incline y, and the direction parallel to the incline x.

If it helps, rotate the incline so that the surface of the incline is flat.

Use trigonometry to determine the components of the weight force. The y component of the weight is:

Wᵧ = mg cos θ

Wᵧ = (1150 kg) (9.8 m/s²) (cos 8.70°)

Wᵧ = 11,100 N

7 0

1 year ago

Help me please to do those problems...

Stells [14]

Keeping in mind that:

- Distance is the length of the total path covered by the person, regardless of the directions of the different parts of motion

- Displacement is just the distance in a straight line between the final point and the initial point

Let's apply these concepts to solve the different parts of the problem:

1. 120 m, 80 m west

The total distance is the sum of the length of the different paths:

distance = 100 + 20 = 120 m

To find the displacement, we need to find the distance between the starting point and the ending point. Assuming the starting point as

x = 0

Michelle moved 100 m westward and 20 m eastward, so the ending point is at

ending point = 100 - 20 = 80 m (westward)

So, the displacement is

displacement = 80 - 0 = 80 m (west)

2. 6400 m, 0

The distance is equal to the length of the track multiplied by the number of laps, so:

$distance = 1600 \cdot 4 = 6400 m$

The track is circular, and the runner completes exactly 4 laps: it means that at the end of the motion, the runner is at her starting point. So ending point and starting point coincide, and so the displacement

displacement = ending point - starting point = 0

3. 40 m; 10 m forward

The total distance is just the sum of the lengths of the different parts of the motion, so:

distance = 20 + 5 + 10 + 5 = 40 m

The find the displacement, we need to assign signs to every direction:

Forward --> positive along the forward-backward direction

Backard --> negative along the forward-backward direction

Right --> positive direction along the right-left direction

Left --> negative direction along the right-left direction

Along the forward-backward direction, the displacement is:

20 m forward and 10 m backward, so 20 - 10 = 10 (forward)

Along the left right direction, the displacement is:

5 m right and 5 m left, so 5 - 5 = 0

So the net displacement is 10 m (forward)

4. 16 km; 4 km south

Again, t total distance is the sum of the lengths of the different parts of the motion, so:

distance = 2 + 4 + 6 + 4 = 16 km

To find the displacement, we assign signs to every direction:

North --> positive along the north-south direction

south --> negative along the north-south direction

east --> positive direction along the east-west direction

west --> negative direction along the east-west direction

Along the north-south direction, the displacement is:

2 km north and 6 km south, so 2 - 6 = -4 km (4 km south)

Along the east-west direction, the displacement is:

4 km east and 4 km west, so 4 - 4 = 0

So the net displacement is 4 km south

5. 37.7 m; 24 m

The distance covered by the skater is the length of half circumference, so given the radius

r = 12 m

The distance is

$distance= \pi r = \pi \cdot (12)=37.7 m$

The displacement is the distance in a straight line between the ending position and the starting position. Since the skater ends her motion halfway around the circle, the distance between the initial and final point is equal to the diameter of the circle (two times the radius). So,

$displacement = 2r = 2 \cdot 12 = 24 m$

3 0

3 years ago

What part of the atom takes up most of the atoms space?

Ahat [919]

Answer:

The electrons

Explanation:

3 0

3 years ago

Read 2 more answers

if a person can jump maximum along distance of 3m ,on the earth how far could be jump on the moon where acceleration due to grav

allochka39001 [22]

Answer:

The person can jump 48 m on the Moon

Explanation:

The question parameters are;

The maximum long jump distance of a person on Earth, $R_{max}$ = 3 m

The acceleration due to gravity on the Moon = 1 ÷ 16 of that on Earth

The distance the person can jump on the Moon is given as follows;

A person performing a jump across an horizontal distance on Earth (under gravitational force) follows the path of the motion of a projectile

The horizontal range, $R_{max}$ , of a projectile motion is found by using the following formula

$R_{max} = \dfrac{u^2}{g}$

Where;

g = The acceleration due to gravity = 9.8 m/s²

Therefore, we have;

$R_{max} = 3 \, m = \dfrac{u^2}{9.8 \, m/s^2 }$

u² = 3 m × 9.8 m/s² = 29.4 m²/s²

Therefore, on the Moon, we have;

The acceleration due to gravity on the Moon, $g_{Moon}$ = 1/16 × g

∴ $g_{Moon}$ = 1/16 × g = 1/16 × 9.8 m/s² ≈ 0.6125 m/s²

$R_{max \ Moon} = \dfrac{u^2}{g_{Moon}} = \dfrac{29.4 \ m^2/s^2}{0.6125 \, m/s^2 } \approx 48 \, m$

The maximum distance the person can jump on the Moon with the same velocity which was used on Earth is $R_{max \ Moon}$ ≈ 48 m

8 0

3 years ago

Singing that is off-pitch by more than about 1% sounds bad. How fast would a singer have to be moving relative to the rest of a

Nina [5.8K]

Answer:

-3.396 m/s or 3.465 m/s

Explanation:

v = Speed of sound in air = 343 m/s

$v_s$ = Relative speed of the singer

f = Observed frequency

f' = Actual frequency

1% change can mean $f=1.01f'$

From the Doppler effect equation we have

$f=f'\dfrac{v}{v+v_s}\\\Rightarrow 1.01f'=f'\dfrac{v}{v+v_s}\\\Rightarrow 1.01=\dfrac{343}{343+v_s}\\\Rightarrow v_s=\dfrac{343}{1.01}-343\\\Rightarrow v_s=-3.396\ m/s$

The velocity is -3.396 m/s

when $f=0.99f'$

$f=f'\dfrac{v}{v+v_s}\\\Rightarrow 0.99f'=f'\dfrac{v}{v+v_s}\\\Rightarrow 0.99=\dfrac{343}{343+v_s}\\\Rightarrow v_s=\dfrac{343}{0.99}-343\\\Rightarrow v_s=3.46464646465\ m/s$

The velocity is 3.465 m/s

3 0

3 years ago