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kherson [118]
3 years ago
14

Define an element and give 5 examples of elements that are important to life .

Physics
1 answer:
fenix001 [56]3 years ago
8 0

Answer:

A pure substance consisting only of atoms with the same number of protons in their nuclei-these appear on the periodic table

Oxygen

Hydrogen

Carbon

Sulfur

Phosphate

Nitrogen

Magnesium

Calcium

Potassium

Chlorine

(I know that these are more examples than needed, but you can use any)

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Identify whether the following statements describe a change in acceleration. Explain your response.
svetlana [45]
They describe a constant rate of acceleration. As in they don't increase at all they stay at a constant rate. The car though stopped because of a stop sign.
4 0
3 years ago
Read 2 more answers
8. You push downward on a box with a force of 750 N. The box is on a flat horizontal surface for which the coefficient of static
natima [27]

Answer:

The mass of the heaviest box you will be able to move with this applied force = 61.4 kg

Explanation:

From the diagram attached, the forces acting on the box include the weight of the box, applied force on the box, normal reaction of the surface on the box and the Frictional force in the opposite direction to the applied force.

For the box to be able to move, the applied force must have a horizontal component that at least matches the Frictional force between the box and the surface. This is the force balance in the horizontal direction.

Resolving the applied force into horizontal and vertical components,

Fₓ = 750 cos 25° = 679.73 N

Fᵧ = 750 sin 25° = 316.96 N

Doing a force balance in the vertical axis,

N = (mg + 316.96)

Frictional force = μN = μ (mg + 316.96)

μ = 0.74, g = 9.8 m/s²

Frictional force = Fᵧ

μ (mg + 316.96) = 679.73

0.74(9.8m + 316.96) = 679.73

7.252m + 234.5504 = 679.73

7.252m = 679.73 - 234.5504 = 445.1796

m = (445.1796/7.252)

m = 61.4 kg

Hope this Helps!!!

6 0
3 years ago
What is the relationship between distance of the objects and the gravitational force between them? (Another way of asking: What
Nadya [2.5K]

Explanation:

increase in the force of gravity

6 0
3 years ago
a spotted lizard runs at 3m/s at top speed. a girl wants to catch the lizard to keep as a pet. where should the girl place her c
Ann [662]

The acceleration due to earth's gravity is -9.8 m/s [dn] I thought... I'm assuming this is a projectile motion question asking for the range.

Break each kinematic quantity into their x and y components.

          x           y

v₁ = 3 m/s        0

v₂ = 3 m/s        ?

Δd = ?            -1.5

Δt  = ?              ?

a = 0           -10 m/s²

So the variable we are trying to find is Δdx (x component of displacement). We need to use a kinematic equation to do so. However, we obviously don't have enough given to find Δdx. This means we need to find something first, something we can use. How about Δt? Δt can be applied to both the x and y components. We have enough information in the y component list to find Δt. We can use this formula and solve for Δt.

Δdy = v₁y ( Δt ) + 1/2 ( ay ) ( Δt )²

Δdy = 1/2 ( ay ) ( Δt )²   <- the first term cancels out since v₁y = 0.

2Δdy = ay ( Δt )²

2Δdy / ay = ( Δt )²

√ 2Δdy / ay = Δt

√ 2(-1.5 m/s) / -9.8 m/s² = Δt

√ -3.0 m/s / -9.8 m/s² = Δt

√ 0.3<u>0</u>6122449 s² = Δt

0.5<u>5</u>32833352 s = Δt

Now, we can use this newly found quantity to solve for Δdx using the x component values using the appropriate kinematic equation.

Δdx = ( v₁x + v₂x / 2) ( Δt )

Δdx = ( ( 3.0 m/s + 3.0 m/s ) / 2 ) ( 0.5<u>5</u>32833352 s )

Δdx = ( 6.0 m/s / 2 ) ( 0.5<u>5</u>32833352 s )

Δdx = ( 3.0 m / s )( 0.5<u>5</u>32833352 s )

Δdx = 1.<u>6</u>59850006 m

Therefore, the girl should place her cage 1.7 m away from the platform to catch the lizard.

This solution assumes that the acceleration due to gravity is -10 m/s² [dn] and not -9.8 m/s² [dn]. If you need -9.8 m/s² [dn], then just substitute it into my solution. This was a pain to type lol





6 0
4 years ago
A 26.0 g ball is fired horizontally with initial speed v0 toward a 110 g ball that is hanging motionless from a 1.10 m -long str
mart [117]

Answer:

7.3 ms^{-1}

Explanation:

Consider the motion of the ball attached to string.

In triangle ABD

Cos50 = \frac{AB}{AD} \\Cos50 = \frac{AB}{L}\\AB = L Cos50

height gained by the ball is given as

h = BC = AC - AD \\h = L - L Cos50\\h = 1.10 - 1.10 Cos50\\h = 0.393 m

M  = mass of the ball attached to string = 110 g

V = speed of the ball attached to string just after collision

Using conservation of energy

Potential energy gained = Kinetic energy lost

Mgh = (0.5) M V^{2} \\V = sqrt(2gh)\\V = sqrt(2(9.8)(0.393))\\V = 2.8 ms^{-1}

Consider the collision between the two balls

m  = mass of the ball fired = 26 g

v_{o} = initial velocity of ball fired before collision = ?

v_{f} = final velocity of ball fired after collision = ?

using conservation of momentum

m v_{o} = MV + m v_{f}\\26 v_{o} = (110)(2.8) + 26 v_{f}\\v_{f} = v_{o} - 11.85

Using conservation of kinetic energy

m v_{o}^{2} = MV^{2} + m v_{f}^{2} \\26 v_{o}^{2} = 110 (2.8)^{2} + 26 (v_{o} - 11.85)^{2} \\v_{o} = 7.3 ms^{-1}

3 0
3 years ago
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