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kati45 [8]
3 years ago
7

Which statement best describes the atoms of elements that form compounds by covalent bonding?

Physics
1 answer:
k0ka [10]3 years ago
4 0

Answer:

they share electrons between them.

Explanation:

taking the test rn lol i think its right

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A nurse pushes a cart by exerting a force on the handle at a downward angle of 36 degrees below the horizontal. the loaded cart
Troyanec [42]

Since it was stated that it must move at constant velocity, so the only force it must overpower is the frictional force.

So the equation is:

F cos θ = Ff

F cos 36 = 65 N

F = 80.34 N

 

<span>So the nurse must exert 80.34 N of force</span>

4 0
3 years ago
A bowling ball moving with a velocity of 5V to the right collides elastically with a beach ball moving at a velocity 2V to the l
katen-ka-za [31]

Answer:

v'_2=3V

Explanation:

From the question we are told that:

Bowling ball Speed v_1=5 m/s

Beach ball Speed v_2=2 m/s

Let The Mass be equal i.e

 M_1=M_2

Therefore

Generally the equation for Velocity of beach ball after collision v'_2 is mathematically given by

Since Velocity is Vector Quantity

Therefore

 v'_2=v_1-v_2

 v'_2=5-2

 v'_2=3V

3 0
3 years ago
Find the velocity, acceleration, and speed of a particle with the given position function. r(t = t2i 6tj 4 ln t k
artcher [175]
1st derivative gives velocity;
d r(t)/ dt = 2t i + 6 j + 4/t k

2nd derivative gives acceleration;
d^2 r(t)/ dt^2 = 2 i - 4/ t^2

Speed ;
Square root of (4 t^2 + 36 + 16/ t^2)

For a given time, like 2 seconds, t will be 2. And answer of speed will be scalar.
6 0
3 years ago
A 60 kg acrobat is in the middle of a 10 m long tightrope. The center of the rope dropped 30 cm in relation to the ends that are
Zigmanuir [339]

Answer:

The tension in each half of the rope, is approximately 4,908.8 N

Explanation:

The mass of the acrobat, m = 60 kg

The length of the rope, l = 10 m

The extent by which the center dropped = 30 cm = 0.3 m

Let, 'T' represent the tension in each half of the rope

Weight, W = Mass, m × The acceleration due to gravity, g

∴ W = m × g

The acceleration due to gravity, g ≈ 9.8 m/s²

∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N

The angle the dropped rope makes with the horizontal, θ is given as follows;

θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°

At equilibrium, the sum of vertical forces, \Sigma F_y = 0

The vertical component of the tension, T_y, in each half of the rope is given as follows;

T_y = T × sin(θ)

∴ \Sigma F_y = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)

Plugging in the values, with θ = arctan(0.06) for accuracy, we get;

588 N + 2 × T × sin(arctan(0.06) = 0

∴ 2 × -T × sin(arctan(0.06) = 588 N

-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N

The tension in each half of the rope, T ≈ 4,908.8 N.

4 0
3 years ago
The equation v=F^aL^M^-c were shows the relationship between velocity of the waves tensile force in the string length, L and mas
Travka [436]
I literally looked everywhere for the answer, and I still found nothing. I hope you get it right. Sorry.
8 0
3 years ago
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