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almond37 [142]
3 years ago
12

Can instantaneous velocity ever be negative?

Physics
2 answers:
kirill115 [55]3 years ago
6 0

Answer:

yes, explained

Explanation:

The instantaneous velocity of an object can be negative. Even though the average velocity of the object in an entire trip is positive, the instantaneous velocity can be negative if the object is moving in negative direction. This direction depends upon the observer. If he considers direction away from him in the right as positive then the direction towards him in the left is negative.

SSSSS [86.1K]3 years ago
4 0
Instantaneous velocity in definition is the average velocity in given particular moment time. Hence, it underlies the context of each velocity at each occupied time and space at that particular moment where you are. Moreover, talking about velocity is vector quantity, which means it both has magnitude and direction. Furthermore, this positive and negative sign attributed to their number means only the specific direction in which the object is going. For example in analogy to average velocity, the initial position of the object at the moment is 2 m/s to the right then the object suddenly changes his position to 3 m/s to the left as his final position. Which is, the exact opposite of the former direction then indicates that the final position is negative, contrary to the initial direction. Just remember that the negative values connoted on the numbers can indicate the opposite direction of either the initial position or an object is downward.



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A pencil has a density of 0.875 g/ml. It has a volume of 4.0 ml. Find the mass
loris [4]

Answer:

To find the mass using density and volume we just multiply them against each other which causes ml to cancel and just leaves us with grams which represents how much the item weights.

mass=density*volume

mass=0.875\frac{g}{ml}*4.0\ ml

mass=3.5\ g

Therefore, our final answer is that our pencil weight 3.5 grams

<u><em>Hope this helps!  Let me know if you have any questions</em></u>

3 0
2 years ago
Larry drops a 5kg ball off of a building. The ball hits the ground 4.7s later. How tall is the building?
musickatia [10]

Explanation:

Given:

v₀ = 0 m/s

a = 9.8 m/s²

t = 4.7 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (4.7 s) + ½ (9.8 m/s²) (4.7 s)²

Δy ≈ 110 m

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3 years ago
SO MANY POINTS! WILL MARK BRAINIEST!!!! PLZ BE FASSTT!
lora16 [44]

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4 0
3 years ago
You toss a rock up vertically at an initial speed of 39 feet per second and release it at an initial height of 6 feet. The rock
3241004551 [841]

Answer:

2.583 s, 29.77 ft and 1.219 s

Explanation:

Using equation of motion and taken the motion upward as positive, also a = g ( acceleration due to gravity) = - 32 fts⁻², V= 39 fts⁻¹ V₁ is final velocity, y is the distance in ft from the ground

H = 6 ft, the height from which it is tossed

V₁ = V + gt = V - gt

at maximum height the body came to rest momentarily V₁ = 0

0 = V - gt

-V = -gt

- 39 / -32 = t

t time to reach maximum height = 1.219 s

To Maximum height reached can be calculated with the formula

V₁² = V² + 2g( y - H) where H is the initial height reached by the tossed rock

where V₁ is the final velocity at maximum height which = 0

0 = V² - 2g(y-H) where y is the distance traveled from the ground

-V² = -2g(y-H)

₋V² / -2g = y-H

(V²/2g) + H = y in ft

(39² / (2 × 32)) + 6

y = 29.77 ft

The total time it will be in air can be calculated with the formula below

y = H + Vt - 0.5gt² from y-H = ut + 0.5at²

0.5gt² - Vt - H = 0 since the body returned to the ground ( y = 0)

0.5gt² - Vt - H = 0

using quadratic formula

- (-V)² ± √ ((-V²) - 4 × 0.5g × -H) / (2 × 0.5 × g)

(V ± √ (V² + 2gH)) ÷ g

substitute the values into the expression

t = (39 + √(39² + (2×-32× 6)))/ 32 or (39 - √ (39² + (2 × -32×6))/ 32

t = (39 + √(1521 +384))/32 = (39 + √1905) / 32  = 2.583 s

t = (39 - √1905) / 32 =  -0.15 s

The will remain in air (V ± √ (V² + 2gH)) / g seconds. It will reach a maximum height of (V²/2g) + H feet after V/g seconds

8 0
3 years ago
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goldfiish [28.3K]

Answer:

Distance from start point is 72.5km

Explanation:

The attached Figure shows the plane trajectories from start point (0,0) to (x1,y1) (d1=40km), then going from (x1,y1) to (x2,y2) (d2=56km), then from (x2,y2) to (x3,y3) (d3=100). Taking into account the angles and triangles formed (shown in the Figure), it can be said:

x1=d1*cos(60), y1=d1*sin(60)\\\\ x2=x1 , y2=y1-d2\\\\ x3=x2-d3*cos(30) , y3=y2+d3*sin(30)

Using the Pitagoras theorem, the distance from (x3,y3) to the start point can be calculated as:

d=\sqrt{x3^{2} +y3^{2} }

Replacing the given values in the equations, the distance is calculated.

4 0
3 years ago
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