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Mkey [24]
3 years ago
10

Derek owns a farm that produces soybeans, avocados, and pecans. What is the word used to describe these items?

Physics
2 answers:
Novosadov [1.4K]3 years ago
7 0

Answer:

crops

Explanation:

dairy are milk products and those aren’t

live stock is domesticated farm animals raised to produce resources

non-perishables are canned food and dried food. they can last a long time. they don’t need to be refrigerated or

Likurg_2 [28]3 years ago
7 0
A: crops
they are not any of the other answer choices
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Describe all the ways that newtons laws can apply in a car crash
Harman [31]
Newtons first law - Objects in the car at rest (The human) will remain at rest unless affected by an unbalanced force. Well the unbalanced force would be the crash and this would set the human in motion and they would ether fly out the car if not wearing a seat belt or if wearing one they would get bad whip lash

Newtons second law - With more mass requires more force, so since the human is pretty light or even if heavy in a big crash there will be so much more from it that this will send the human flying.

Newtons 3rd law - Objects A puts force onto objects b and object b excretes the same amount of force back onto object a, so in a crash the human would hit the car hard and the car would excrete the same amount of force back on the human which would really damage him/her
7 0
3 years ago
What si unit measures speed ?
dlinn [17]
Hmm doesnt soujd familiar
4 0
3 years ago
A ball is thrown downward at 12 m/s from a windowsill 35 m above the ground. At the same time, another ball is thrown upward at
wariber [46]

Answer:

The second ball lands 1.5 s after the first ball.

Explanation:

Given;

initial velocity of the ball, u = 12 m/s

height of fall, h = 35 m

initial velocity of the second, v = 12 m/s

Time taken for the first ball to land;

t = \sqrt{\frac{2h}{g} }\\\\t =\sqrt{ \frac{2*35}{9.8}}\\\\t = 2.67 \ s

determine the maximum height reached by the second ball;

v² = u² -2gh

at maximum height, the final velocity, v = 0

0 = 12² - (2 x 9.8)h

19.6h = 144

h = 144 / 19.6

h = 7.35 m

time to reach this height;

t_1 = \sqrt{\frac{2h}{g} }\\\\t_1 =  \sqrt{\frac{2*7.35}{9.8}}\\\\t_1 = 1.23 \ s

Total height above the ground to be traveled by the second ball is given as;

= 7.35 m + 35m

= 42.35 m

Time taken for the second ball to fall from this height;

t_2 = \sqrt{\frac{2h}{g} }\\\\t_2 = \sqrt{\frac{2*42.35}{9.8} }\\\\t_2 = 2.94 \ s

total time spent in air by the second ball;

T = t₁ + t₂

T = 1.23 s + 2.94 s

T = 4.17 s

Time taken for the second ball to land after the first ball is given by;

t = 4.17 s -  2.67 s

T = 1.5 s

Therefore, the second ball lands 1.5 s after the first ball.

4 0
3 years ago
Two small charged spheres are located on the y-axis. One is at y = 1.00 m, the other is at y = −1.00 m, and they both have a cha
yuradex [85]

Answer:

(a) 23.946 kV

(b) -0.077 J

Explanation:

(a) The electric potential is given by the following formula:

V=k\frac{q_1}{r_1}+k\frac{q_2}{r_2}   (1)

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

q1 = q2 = 1.60*10^{-6}C

r1 and r2 are the distance from the charges to the point in which electric potential is evaluated.

Firs, you calculate the distance r1 and r2 by taking into account the position of the charges

r_1=\sqrt{(1.00)^2+(0.670)^2}m=1.20m\\\\r_2=\sqrt{(-1.00)^2+(0.670)^2}m=1.20m

Next, you replace the values of the parameters to calculate V:

V=(8.98*10^9Nm^2/C^2)\frac{1.6*10^{-6}C}{1.20m}+(8.98*10^9Nm^2/C^2)\frac{1.6*10^{-6}C}{1.20m}\\\\V=23946.66\ V=23.946\ kV

(b) The potential electric energy is given by:

U_T=U_{1,2}+U_{1,3}+U_{2,3}\\\\U_T=k\frac{q_1q_2}{r_{1,2}}+k\frac{q_1q_3}{r_{1,3}}+k\frac{q_2q_3}{r_{2,3}}\\\\r_{1,2}=2.00m\\\\r_{1,3}=1.20m\\\\r_{2,3}=1.20m\\\\U_T=(8.98*10^9)[\frac{(1.6*10^{-6})^2}{2.00m}+\frac{(1.6*10^{-6})(-3.70*10^{-6})}{1.20}+\frac{(1.6*10^{-6})(-3.70*10^{-6})}{1.20}]J\\\\U_T=-0.077J

5 0
3 years ago
A certain instant camera determines the distance to the subject by sending out a sound wave and measuring the time needed for th
valentinak56 [21]
The total distance covered by the sound wave is twice the distance between the camera and the subject (because the wave has to reach the subject and then travel back to the camera), so 2L, where L=3.42 m. The speed of sound is v=343 m/s. It is a uniform linear motion, so we can use the basic relationship between space (S), time (t) and velocity (v) to find the time the wave needs to return to the camera:
t= \frac{S}{v}= \frac{2L}{v}= \frac{2\cdot 3.42 m}{343 m/s}=0.020 s
6 0
3 years ago
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