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rewona [7]
3 years ago
8

Plz help me to find my science matching... ​

Physics
1 answer:
s2008m [1.1K]3 years ago
5 0

Answer:

fibrous =potato

taproot =radish

stilt =maize and sugar cane

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Why is it not necessary for radio telescope surfaces to be as smooth as a mirror?
Law Incorporation [45]
It doesn't on account of radio waves are longer than optical waves. Radio waves are a sort of electromagnetic radiation with wavelengths in the electromagnetic range longer than infrared light. These long waves are in the radio locale of the electromagnetic range.
8 0
3 years ago
Help please I do not understand
ahrayia [7]
What don’t you understand? If you haven’t uploaded anything
3 0
3 years ago
Emmit is lifting a box vertically which forces are necessary for calculating the total force
zavuch27 [327]

Answer:

FG and FP

Explanation:

Gravitational Force(FG) because the box is being pulled down and a resistance force pushing up because he is pushing(FP) up on the box.

I hope this helps!

4 0
3 years ago
A girl exerts a horizontal force of 109 N on a crate with a mass of 31.2 kg. HINT (a) If the crate doesn't move, what's the magn
vesna_86 [32]

Answer:

(a) Magnitude of static friction force is 109 N

(b) Minimum possible value of static friction is 0.356

Solution:

As per the question;

Horizontal force exerted  by the girl, F = 109 N

Mass of the crate, m = 31.2 kg

Now,

(a) To calculate the magnitude of static friction force:

Since, the crate is at rest, the forces on the crate are balanced and thus the horizontal force is equal to the frictional force, f:

F = f = 109 N

(b) The maximum possible force of friction between the floor and the crate is given by:

f_{m} = \mu_{s}N

where

N = Normal reaction = mg

Thus

f_{m} = \mu_{s}mg

For the crate to remain at rest, The force exerted on the crate must be less than or equal to the maximum force of friction.

f\leq f_{m}

f \leq \mu_{s}mg

109 \leq \mu_{s}\times 31.2\times 9.8

\mu_{s}\geq 0.356

7 0
3 years ago
A 0.58 kg mass is moving horizontally with a speed of 6.0 m/s when it strikes a vertical wall. The mass rebounds with a speed of
Sunny_sXe [5.5K]

Answer:

5.8\; {\rm kg\cdot m \cdot s^{-1}}.

Explanation:

If the mass of an object is m and the velocity of that object is v, the linear momentum of that object would be m\, v.

Assume that the initial velocity of the mass is positive (6.0\; {\rm m\cdot s^{-1}}.) However, the direction of the velocity is reversed after the impact. Thus, the sign of the new velocity of the object would be negative- the opposite of that of the initial velocity. The new velocity would be (-4.0\; {\rm m\cdot s^{-1}}).

Thus, the change in the velocity of the mass would be:

\begin{aligned}& (\text{Change in Velocity}) \\ =\; & (\text{Final Velocity}) - (\text{Initial Velocity}) \\ =\; & (-4.0\; {\rm m\cdot s^{-1}}) - (6.0\; {\rm m\cdot s^{-1}}) \\ =\; & (-10\; {\rm m\cdot s^{-1})\end{aligned}.

The change in the linear momentum of the mass would be:

\begin{aligned} & \text{change in momentum} \\ =\; & (\text{mass}) \times (\text{change in velocity}) \\ =\; & 0.58\; {\rm kg} \times (-10\; {\rm m\cdot s^{-1}}) \\  =\; & (-5.8\; {\rm kg \cdot m \cdot s^{-1}})\end{aligned}.

Thus, the magnitude of the change of the linear momentum would be 5.8\; {\rm kg \cdot m \cdot s^{-1}}.

7 0
2 years ago
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