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3 years ago

Three plant like organisms which cannot produce their own food are

1 answer:

8 0

I can only offer you one which is mushroom. They're fungus' and they can't produce their own food. They get their food from decaying and dead animals and absorb it through their roots.

You might be interested in

On a horizontal surface is located

Ierofanga [76]

By Newton's second law, the net vertical force acting on the object is 0, so that

n - w = 0

where n = magnitude of the normal force of the surface pushing up on the object, and w = weight of the object. Hence n = w = mg = 196 N, where m = 20 kg and g = 9.80 m/s².

The force of static friction exerts up to 80 N on the object, since that's the minimum required force needed to get it moving, which means the coefficient of static friction µ is such that

80 N = µ (196 N) → µ = (80 N)/(196 N) ≈ 0.408

Moving at constant speed, there is a kinetic friction force of 40 N opposing the object's motion, so that the coefficient of kinetic friction ν is

40 N = ν (196 N) → ν = (40 N)/(196 N) ≈ 0.204

And so the closest answer is C.

(Note: µ and ν are the Greek letters mu and nu)

3 0

3 years ago

Police radar equipment is used to detect the speed of objects. In one trial, the radar equipment records a stationary tree as tr

Brut [27]

Answer:

reliability

accuracy

Explanation:

If a reading of a measurement is consistently the same then the measurement is reliable.

If a reading of measurement is close the actual value of the measurement then the reading is accurate.

Here, a stationary tree shows reading 6 mph once and 0 mph another instant. So, neither the reading of a measurement is consistent not the reading of measurement is close the actual value.

Hence, the radar has problems in its reliability and accuracy

8 0

3 years ago

A brick lands 10.1 m from the base of a building. If it was given an initial velocity of 8.6 m/s [61º above the horizontal], how

Montano1993 [528]

<h2>Answer: 10.52m</h2><h2 />

First, we have to establish the reference system. Let's assume that the building is on the negative y-axis and that the brick was thrown at the origin (see figure attached).

According to this, the initial velocity $V_{o}$ has two components, because the brick was thrown at an angle $\alpha=61\º$ :

$V_{ox}=V_{o}cos\alpha$ (1)

$V_{ox}=8.6\frac{m}{s}cos(61\º)=4.169\frac{m}{s}$ (2)

$V_{oy}=V_{o}sin\alpha$ (3)

$V_{oy}=8.6\frac{m}{s}sin(61\º)=7.521\frac{m}{s}$ (4)

As this is a projectile motion, we have two principal equations related:

$X=V_{ox}.t$ (5)

Where:

$X=10.1m$ is the distance where the brick landed

$t$ is the time in seconds

If we already know $X$ and $V_{ox}$ , we have to find the time (we will need it for the following equation):

$t= \frac{X}{ V_{ox}}$ (6)

$t=2.42s$ (7)

$-y=V_{oy}.t+\frac{1}{2}g.t^{2}$ (8)

Where:

$y$ is the height of the building (in this case it has a negative sign because of the reference system we chose)

$g=-9.8\frac{m}{s^{2}}$ is the acceleration due gravity

Substituting the known values, including the time we found on equation (7) in equation (8), we will find the height of the building:

$-y=(7.521\frac{m}{s})(2.42s)+\frac{1}{2}(-9.8\frac{m}{s^{2}}).(2.42s)^{2}$ (9)

$-y=-10.52m$ (10)

Multiplying by -1 each side of the equation:

$y=10.52m$ >>>>This is the height of the building

3 0

3 years ago

PLEASE HELP AND HURRY

Ahat [919]

Answer:

Explanation:

6 0

3 years ago

Read 2 more answers

31. If you threw a baseball straight out at 45 m/s from a height of 1.5 meters (A) how long would it be in the air? B) How far o

coldgirl [10]

Answer:

A) t = 0.55 s

B) x = 24.8 m

Explanation:

A) We can find the time at which the ball will be in the air using the following equation:

$y_{f} = y_{0} + v_{0y}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$ is the final height= 0

$y_{0}$ is the initial height= 1.5 m

$v_{0y}$ is the component of the initial speed in the vertical direction = 0 m/s

t: is the time =?

g: is the gravity = 9.81 m/s²

$0 = 1.5 m - \frac{1}{2}9.81 m/s^{2}t^{2}$

By solving the above equation for t we have:

$t = \sqrt{\frac{2*1.5 m}{9.81 m/s^{2}}} = 0.55 s$

Hence, the ball will stay 0.55 seconds in the air.

B) We can find the distance traveled by the ball as follows:

$x_{f} = x_{0} + v_{0x}t + \frac{1}{2}at^{2}$

Where:

a: is the acceleration in the horizontal direction = 0

$x_{f}$ is the final position =?

$x_{0}$ is the initial position = 0

$v_{0x}$ is the component of the initial speed in the horizontal direction = 45 m/s

$x_{f} = x_{0} + v_{0x}t + \frac{1}{2}at^{2}$

$x_{f} = 0 + 45 m/s*0.55 s + 0 = 24.8 m$

Therefore, the ball will travel 24.8 meters.

I hope it helps you!

3 0

2 years ago