All categories

2 years ago

Producer: (plants) (produces its own food) 1. The common name of my organism is: ______________

Physics

2 answers:

Lelu [443]2 years ago

6 0

1. Root
2. Food

This is the answer hope u like it and plz mark me as brainlyist.

Mama L [17]2 years ago

4 0

Answer: 1. Plant

2. Glucose

Explanation:

You might be interested in

How can having a mentor help individuals achieve their goals? A. Mentors can increase accountability. B. Mentors can increase mo

lana [24]

Your best bet is to go with D.

6 0

2 years ago

Read 2 more answers

Ask Your Teacher Cam Newton of the Carolina Panthers throws a perfect football spiral at 6.9 rev/s. The radius of a pro football

faltersainse [42]

Answer:

$a=159.32\ m/s^2$

Explanation:

It is given that,

Angular speed of the football spiral, $\omega=6.9\ rev/s=43.35\ rad/s$

Radius of a pro football, r = 8.5 cm = 0.085 m

The velocity is given by :

$v=r\omega$

$v=0.085\times 43.35$

v = 3.68 m/s

The centripetal acceleration is given by :

$a=\dfrac{v^2}{r}$

$a=\dfrac{(3.68)^2}{0.085}$

$a=159.32\ m/s^2$

So, the centripetal acceleration of the laces on the football is $159.32\ m/s^2$ . Hence, this is the required solution.

6 0

3 years ago

A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the

alex41 [277]

Answer:

(a) $v_1 = a_1t_1 = 1.76 t_1$

(b) It won't hit

Explanation:

(a) the car velocity is the initial velocity (at rest so 0) plus product of acceleration and time t1

$v_1 = v_0 + a_1t_1 = 0 +1.76t_1 = 1.76t_1$

(b) The velocity of the car before the driver begins braking is

$v_1 = 1.76*20 = 35.2m/s$

The driver brakes hard and come to rest for t2 = 5s. This means the deceleration of the driver during braking process is

$a_2 = \frac{\Delta v_2}{\Delta t_2} = \frac{v_2 - v_1}{t_2} = \frac{0 - 35.2}{5} = -7.04 m/s^2$

We can use the following equation of motion to calculate how far the car has travel since braking to stop

$s_2 = v_1t_2 + a_2t_2^2/2$

$s_2 = 35.2*5 - 7.04*5^2/2 = 88 m$

Also the distance from start to where the driver starts braking is

$s_1 = a_1t_1^2/2 = 1.76*20^2/2 = 352$

So the total distance from rest to stop is 352 + 88 = 440 m < 550 m so the car won't hit the limb

8 0

2 years ago

Please answer D in the image with an explanation

puteri [66]

Answer:

The force is 274 N.

Explanation:

In figure 2:

(d) Let the tension in the string is T.

According to the Newton's second law,

Net force = mass x acceleration

Apply for 200N.

$T - 200 sin 35 =\frac{200}{9.8}\times a \\T - 114.7 = 20.4 a..... (1)\\220 - T = \frac{220}{9.8}\times a\\220 - T = 22.45 a..... (2)\\Adding both the equations\\334.7 = 42.85 aa =7.81 m/s^{2}$

Now put in (1)

T - 114.7 = 20.4 x 7.81

T = 274 N

4 0

2 years ago

True or False: The basketball should be dribbled below the waist.

zimovet [89]

True if you have proper stance and use your body the right way then the ball will be below your waist to allow for more control.

7 0

2 years ago