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maxonik [38]
2 years ago
11

Easy one here pls help thanks so much

Physics
2 answers:
Kisachek [45]2 years ago
8 0
The answer is B. Nutrients.
Alexus [3.1K]2 years ago
7 0
The answer is B. Nutrients
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A 0.03 kg golf ball is hit off the tee at a speed of 34 m/s. The golf club was in contact with the ball for 0.003 s. What is the
Liula [17]

Answer:

The average force on ball by the golf club is 340 N.

Explanation:

Given that,

Mass of the golf ball, m = 0.03 kg

Initial speed of the ball, u = 0

Final speed of the ball, v = 34 m/s

Time of contact, \Delta t=0.003\ s

We need to find the average force on ball by the golf club. We know that the rate of change of momentum is equal to the net external force applied such that :

F=\dfrac{\Delta p}{\Delta t}\\\\F=\dfrac{mv-mu}{\Delta t}\\\\F=\dfrac{mv}{\Delta t}\\\\F=\dfrac{0.03\ kg\times 34\ m/s}{0.003\ s}\\\\F=340\ N

So, the average force on ball by the golf club is 340 N.

4 0
3 years ago
Heres a random question just for fun and out. of boredom ok so who started the nasa program? and why? what made him want to do t
Pachacha [2.7K]
Eisenhower started the NASA project to develop technology for military application.
5 0
3 years ago
A crate is pushed horizontally by a horizontal force 527.018 N . Sliding friction resists the motion, and the kinetic coefficien
gulaghasi [49]

Answer:

m= 10 kg a = 52 m / s²

Explanation:

For this problem we must use Newton's second law, let's apply it to each axis

X axis

      F - fr = ma

The equation for the force of friction is

    -fr = miu N

Axis y

     N- W = 0

     N = mg

Let's replace and calculate laceration

     F - miu (mg) = ma

    a = F / m - mi g

    a = 527.018 / m - 0.17 9.8

We must know the mass of the body suppose m = 10 kg

    a = 527.018 / 10 - 1,666

    a = 52 m / s²

5 0
3 years ago
A point charge A of charge +4micro coloumb and another B of -1 micro coloumb are placed at a distance in air 1m apart then the d
andrew11 [14]

Answer:

Explanation:

Given that,

A point charge is placed between two charges

Q1 = 4 μC

Q2 = -1 μC

Distance between the two charges is 1m

We want to find the point when the electric field will be zero.

Electric field can be calculated using

E = kQ/r²

Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.

Then, the magnitude of the electric at point x is zero.

E = kQ1 / r² + kQ2 / r²

0 = kQ1 / x²  - kQ2 / (1-x)²

kQ1 / x² = kQ2 / (1-x)²

Divide through by k

Q1 / x² = Q2 / (1-x)²

4μ / x² = 1μ / (1 - x)²

Divide through by μ

4 / x² = 1 / (1-x)²

Cross multiply

4(1-x)² = x²

4(1-2x+x²) = x²

4 - 8x + 4x² = x²

4x² - 8x + 4 - x² = 0

3x² - 8x + 4 = 0

Check attachment for solution of quadratic equation

We found that,

x = 2m or x = ⅔m

So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.

5 0
3 years ago
What part of the plant takes in carbon dioxide?
murzikaleks [220]

The answer is number 2 stomata.

4 0
3 years ago
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