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Sunny_sXe [5.5K]
3 years ago
15

Listed in the Item Bank are some important labels for sections of the image below. To find out more information about labels, so

me have more details available when you click on them. Drag and drop each label to the corresponding area it identifies in the image.

Physics
1 answer:
Agata [3.3K]3 years ago
6 0

Answer:

to the left the first answer can be "Consumer" and under it can go "plant"

to the right the first answer on the top can be "Carnivore" under it can go "Herbivore" and under it can go "Producer"

I just had this question and its right

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How does the number 10 relate to light, sound, force and motion?
Oksana_A [137]

Answer:

number 10 is the value of acceleration due to gravity.

8 0
3 years ago
Under what conditions will an object experience a gravitational force?
Radda [10]
A) an object with mass > 0 in a gravitational field
b) an object with an electric charge not 0 in an electric field
c) a moving object with an electric charge not 0 in a magnetic field
6 0
3 years ago
a piece of clay with mass m is traveling with an unknown velocity in the positive x-direction, when it collides and sticks to a
joja [24]

Answer:

13 m in the + x direction.

Explanation:

Givens

m1 = m

m2 = 3m

v1= ? = x

v2 = - 3 m/s

v3 = v4 = - 1 m/s

Equation

Let x be the speed of the lighter object before collision.

Momentum before collision = momentum after collision

Solution

mx   - 3m * 3m/s = (m1 + m2)*(-1)

m(x - 3*3m/s) = (m + 3m)(-1)

m(x - 9) = - 4m                                      Cancel the ms

x - 9 = 4                                                 Add 9 to both sides

x = 9 + 4

x = 13 m/s

7 0
3 years ago
Community psychologists are not<br> concerned with worker stress factors.<br> True<br> False
fgiga [73]
It would be false as your answer
8 0
2 years ago
Please help !! ASAP
umka2103 [35]

Answer:

22) 5.6 Kg.m/s

23) 0.067 Kg

24) 1.16 m/s

Explanation:

Momentum is found by multiplying mass and velocity

P=mv where P is momentum, m is mass of object and v is the velocity

22

Given

Mass, m=2 Kg

Rate, v=2.8 m/s

Relationship

P=mv

Solution

P=mv=2*2.8=5.6 Kg.m/s

23

Given

Momentum, p=0.1 Kg.m/s

Rate, v=1.5 m/s

Relationship

P=mv and making m the subject then

m=\frac {P}{v}

Solution

m=\frac {0.1 Kgm.s}{1.5 m/s}\approx 0.067 Kg

24

Given

Momentum, p=4.9\times 10^{8} Kg.m/s

Mass, m=4.23\times10^{8} Kg

Relationship

P=mv and making v the subject then

v=\frac {P}{m}

Solution

m=\frac {4.9\times 10^{8}  Kgm.s}{4.23\times10^{8} }\approx 1.16 m/s

5 0
2 years ago
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