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kirill [66]
3 years ago
7

1. How many grams are in a dekagram?

Physics
2 answers:
tatuchka [14]3 years ago
5 0

Answer:

There are 10 Grams per Dekagram

zepelin [54]3 years ago
4 0

Answer: 10

Explanation:

none just search it up

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A truck loaded with sand accelerates along a highway. The driving force on the truck remains constant. What happens to the accel
pickupchik [31]

Answer:

Acceleration will increase.

Explanation:

The relation between force, mass and acceleration according to the Newton's second law of motion is given as:

F = ma

We are given that the driving force on the truck remains constant, so F is constant here. We can rewrite the above equation as:

a=\frac{F}{m}

Since, F is constant, the acceleration of the truck is inversely proportional to the mass.

There is a hole at the bottom of the truck through which the sand is being lost at a constant rate. Since, the sand is being lost, the overall mass of the truck is being reduced.

Since, the acceleration of the truck is inversely proportional to the mass, the reduced mass will result in an increased acceleration.

So, the acceleration of the truck will increase.

4 0
3 years ago
In a concave mirror parallel rays falling on it convergs at
ella [17]

Answer:

1) In a concave mirror parallel rays falling on it converges at F and 2F.

Explanation:

Spherical mirrors can be used for magnification of images. There are basically two types of spherical mirrors and they are converging mirror and diverging mirrors. The converging mirrors are also termed as concave mirrors and its basic work is to converge or combine light rays coming from a larger distance to a single point. Mostly the light beams falling parallel to the principle axis of the concave mirror will be acting as parallel rays. And when these parallel rays fall on the mirror, the converging point can be the focal point of the mirror.

Thus the location of converging point in concave mirrors will be based on the position or distance of object from the mirror. If the object distance is very far from the twice the focal length distance of mirror, then the converging point will be the focal point or F. And if the object is placed slightly greater than twice the distance of focal point, then the image will be obtained at 2F. But the parallel beams will be converging at F and 2F.

5 0
3 years ago
Which example involves the transformation of chemical energy directly into light energy?
zimovet [89]

burning sodium or magnesium

8 0
3 years ago
What is the rabbit's displacement from t = 0s to 3 s?<br> Answer with two significant digits.
Inessa05 [86]

Answer: i think the answer is 20.0s

Explanation:

3 0
3 years ago
Read 2 more answers
Willie, in a 100.0 m race, initially accelerates uniformly from rest at 2.00 m/s2 until reaching his top speed of 12.0 m/s. He m
Oduvanchick [21]

Answer:

The total time for the race is 11.6 seconds

Explanation:

The parameters given are;

Total distance ran by Willie = 100.0 m

Initial acceleration = 2.00m/s²

Top speed reached with initial acceleration = 12.0 m/s

Point where Willie start to fade and decelerate = 16.0 m from the finish line

Speed with which Willie crosses the finish line = 8.00 m/s

The time and distance covered with the initial acceleration are found using the following equations of motion;

v = u₀ + a·t

v² = u₀² + 2·a·s

Where:

v = Final velocity reached with the initial acceleration = 12.0 m/s

u₀ = Initial velocity at the start of the race = 0 m/s

t = Time during acceleration

a = Initial acceleration = 2.00 m/s²

s = Distance covered during the period of initial acceleration

From, v = u₀ + a·t, we have;

12 = 0 + 2×t

t = 12/2 = 6 seconds

From, v² = u₀² + 2·a·s, we have;

12² = 0² + 2×2×s

144 = 4×s

s = 144/4 =36 meters

Given that the Willie maintained the top speed of 12.0 m/s until he was 16.0 m from the finish line, we have;

Distance covered at top speed = 100 - 36 - 16 = 48 meters

Time, t_t of running at top speed = Distance/velocity = 48/12 = 4 seconds

The deceleration from top speed to crossing the line is found as follows;

v₁² = u₁² + 2·a₁·s₁

Where:

u₁ = v = 12 m/s

v₁ = The speed with which Willie crosses the line = 8.00 m/s

s₁ = Distance covered during decelerating = 16.0 m

a₁ = Deceleration

From which we have;

8² = 12² + 2 × a × 16

64 = 144 + 32·a

64 - 144 = 32·a

32·a = -80

a = -80/32 = -2.5 m/s²

From, v₁ = u₁ + a₁·t₁

Where:

t₁ = Time of deceleration

We have;

8 = 12 + (-2.5)·t₁

t₁ = (8 - 12)/(-2.5) = 1.6 seconds

The total time = t + t_t + t₁ =6 + 4 + 1.6 = 11.6 seconds.

6 0
3 years ago
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