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Fed [463]
3 years ago
11

How does the behavior of oobleck change as you increase the force

Physics
1 answer:
SIZIF [17.4K]3 years ago
8 0
Hope this helps
If u hit oobleck hard it is a solid and when you pick it up and let it fall of your hand it turns into a liquid
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Please help im failing!!!
natta225 [31]
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5 0
3 years ago
While John is traveling along a straight interstate highway, he notices that the mile marker reads 249 km. John travels until he
erastova [34]

Answer:

Explanation:

Displacement can be displayed as a vector, this because it has magnitud and direction. Because of this, we can think John's Resultant Displacement as the join of this two vectors.

The First Vector is from the 249 Km Marker to the 141 Km Marker, which give us a Vector with a Magnitude equals to 108 Km.

The Second Vector goes from 141 Km Marker to the 174 Km Marker, which give us a Vector with a Magnitude equals to 33 Km.

However is important to know the direction for each Vector, we notice that John was traveling on one direction and then he returned. This makes our Vector to have a different direction, and this means difference signs. Difference signs means substraction. So, the Third Vector will be:

Third Vector = 108 Km - 33 Km

Third Vector = 75 Km

8 0
4 years ago
A 0.12 g honeybee acquires a charge of +24pC while flying. The earth's electric field near the surface is typically 100 N/C, dow
shusha [124]

Answer:

150000000

\dfrac{F_e}{F_g}=0.00000203873598369

49050000 N/C

Explanation:

q = Charge = 24 pC

m = Mass of honeybee = 0.12 g

E = Electric field = 100 N/C

g = Acceleration due to gravity = 9.81 m/s²

1\ C=6.25\times 10^{18}\ electrons

Number electrons is

n=24\times 10^{-12}\times 6.25\times 10^{18}\\\Rightarrow n=150000000

The number of electrons added or removed was 150000000

Force is given by

F_e=Eq\\\Rightarrow F_e=100\times 24\times 10^{-12}\\\Rightarrow F_e=2.4\times 10^{-9}\ N

The ratio is

\dfrac{F_e}{F_g}=\dfrac{2.4\times 10^{-9}}{0.12\times 10^{-3}\times 9.81}\\\Rightarrow \dfrac{F_e}{F_g}=0.00000203873598369

The ratio is \dfrac{F_e}{F_g}=0.00000203873598369

Balancing the forces we get

Eq=mg\\\Rightarrow E=\dfrac{mg}{q}\\\Rightarrow E=\dfrac{0.12\times 10^{-3}\times 9.81}{24\times 10^{-12}}\\\Rightarrow E=49050000\ N/C

The electric field required is 49050000 N/C

4 0
4 years ago
Two students are watching a person riding a skateboard up and down a ramp. Each student shares what they think about the energy
avanturin [10]

Answer:

Explanation:

We know that , If the frictional force on a system is zero , then the total energy of a system will be conserved.

By using energy conservation

KE₁ +  U₁ = KE₂ + U₂

KE₁=Kinetic energy at location 1

U₁ =Potential energy at location 1

KE₂=Kinetic energy at location 2

U₂=Potential energy at location 2

Therefore, Raymond is thinking in a right way.

7 0
3 years ago
Find the height or length of these natural wonders in km, m, and cm.A. A cave system with a mapped length of 354 miles.B. A wate
cricket20 [7]

Explanation:

Some standard unit conversion are 1 mile = 1.609344 km, 1 ft.= 30.48 cm,

1 km= 1000 m or 1 m = 0.001 km and 1 m= 100 cm or 1 cm=0.01 m.

Now, use these values to convert the given lengths.

A. length of cave = 5 miles (given)

From standard value 1 mile = 1.609344 km

\Rightarrow 5 miles = \times 1.609344 km= 8.04672 km,

\Rightarrow 5 miles =8.04672 \times 1000 m= 8046.72 m [ as 1 km = 1000 m]

\Rightarrow 5 miles = 8046.72 \times 100 cm=804672 cm

B. Height of the waterfall = 1235 ft. (given)

1 ft.= 30.48 cm [ from standard value]

\Rightarrow 1235.2 ft.= 1235.2\times 30.48 cm=37648.896 cm,

\Rightarrow 1235.2 ft.=37648.896 \times 0.01 m =376.48896 m [ as 1 cm = 0.01 m]

\Rightarrow 1235.2 ft.=376.48896 \times 0.001 km =0.37648896 km [ as 1 m = 0.001 km]

C. Height of the mountain= 21320 ft. (given)

From standard value: 1 ft.= 30.48 cm

\Rightarrow 21320 ft.= 21320 \times 30.48 cm=649833.6 cm,

\Rightarrow 21320 ft.=649833.6 \times 0.01 m =6498.336 m [ as 1 cm = 0.01 m]

\Rightarrow 21320 ft.=6498.336 \times 0.001 km =6.498336 km [ as 1 m = 0.001 km].

D. Depth of canyon =6630 ft.

From standard value: 1 ft.= 30.48 cm

\Rightarrow 6630 ft.= 6630 \times 30.48 cm=202082.4 cm,

\Rightarrow 6630 ft.=202082.4  \times 0.01 m =2020.824  m [ as 1 cm = 0.01 m]

\Rightarrow 6630 ft.=2020.824  \times 0.001 km =2.020824  km [ as 1 m = 0.001 km].

4 0
4 years ago
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