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cricket20 [7]
4 years ago
15

All of the following are very helpful when identifying evidence of past living plants except

Physics
1 answer:
m_a_m_a [10]4 years ago
4 0
JAYLA OTHER KIDS ANSWER OUR QUESTIONS DID YOU KNOW THAT BESTFRIEND
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A kid moving 3.18 m/s jumps on a slip and slide(Uk=0.0751). How far has he slid when his velocity is reduced to 2.00 m/s?
STALIN [3.7K]
F=ma (Newton's second law of motion).

Then,
a = F/m = Uk*mg/m = Uk*g = 0.0751*9.81 = 0.737 m/s^2 (should be negative since a deceleration is expected).

v = Sqrt (u^2-2ad), Where v = final velocity, u=initial velocity, a = acceleration, d = distance moved.

Therefore,

v^2-u^2 = -2ad => d = (v^2-u^2)/-2a = (2^2-3.18^2)/-2*0.737 = 4.15 m
7 0
4 years ago
Drinking water helps your body maintain
8090 [49]

Drinking water helps maintain nearly all processes of the human body and <em>homeostasis. </em>Further, drinking water helps,

  • Digestion
  • Absorption
  • Circulation
  • Saliva and mucus production
  • Thermoregulation/homeostasis
  • Lubrication of joints
  • Boosts integumentary health.
  • Helps in excretory systems
  • Prevents dehydration of cells
  • Reduces chance of hangover from alcoholic drinks.
  • etc,

Succinctly, it is VERY important to drink water! Water is the source of human life and without water, we cannot survive.

6 0
3 years ago
A block slides down a frictionless plane having an inclination of 15.0°. The block starts from rest at the top, and the length o
nalin [4]

Answer: check the pic

Explanation:

8 0
3 years ago
A certain planet has a radius of 4990 km. If, on the surface of that planet, a 95.0 kg object has a weight of 591 N, then what i
aniked [119]

Answer:

3743.489 kg

Explanation:

F_g = 591 N

G = 6.674x10^-11 constant of gravity

m_1 = 95 kg

m_2 = unknown

r = 4990*1000 =

F_g = G[(m_1*m_2)/r^2]

591 N = 6.674x10^-11[(95*m_2)/4990^2]

8.855 = [(95*m_2)/4990^2]

355631.472 = 95*m_2

m_2 = 3743.489 kg

7 0
3 years ago
A small object with mass 1.30 kg is mounted on one end of arod
Julli [10]

Answer:

(a) I_{system} = 1.014\ kg.m^{2}

(b) \tau = 0.0179\ N-m

Solution:

As per the question:

Mass of the object, m = 1.30 kg

Length of the rod, L = 0.780 m

Angular speed, \omega = 5010\ rev/min

Now,

(a) To calculate the rotational inertia of the system about the axis of rotation:

Since, the rod is mass less, the moment of inertia of the rotating system and that of the object about the rotation axis will be equal:

I_{system} = ML^{2} = 1.30\times (0.780)^{2} = 0.791\ kg.m^{2}

(b) To calculate the applied torque required for the system to rotate at constant speed:

Drag Force, F = 2.30\times 10^{- 2}\ N

\tau = FLsin\theta 90 = 2.30\times 10^{- 2}\times 0.780\times 1 = 0.0179\ N-m

3 0
3 years ago
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