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vampirchik [111]
3 years ago
8

A 10 kg bowling ball sits at the top of a 10 m hill and then slides down its icy hillside.

Physics
2 answers:
Nataly_w [17]3 years ago
5 0
We need more informtion
andriy [413]3 years ago
4 0
So is there an answer choice if not the answer would be 100m
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The term used to describe the quantity of matter that a body possesses is:
barxatty [35]
The term used to describe the quantity of matter that a body possesses is mass.
7 0
3 years ago
If the ball shown in the figure lands in 0.5 s, about what height was it thrown from?
Ede4ka [16]

Answer: 1.22 m

Explanation:

The equation of motion in this situation is:

y=y_{o}+V_{oy}t-\frac{g}{2}t^{2} (1)

Where:

y=0 is the final height of the ball

y_{o}=h is the initial height of the ball

V_{oy}=V_{o}sin(0\°)=0 is the vertical component of the initial velocity (assuming the ball was thrown vertically and there is no horizontal velocity)

t=0.5 s is the time at which the ball lands

g=9.8 m/s^{2} is the acceleration due gravity

So, with these conditions the equation is rewritten as:

h=\frac{g}{2}t^{2} (2)

h=\frac{9.8 m/s^{2}}{2}(0.5 s)^{2} (3)

Finally:

h=1.22 m

4 0
3 years ago
A 1.50-m string of weight 0.0125 N is tied to the ceil- ing at its upper end, and the lower end supports a weight W. Ignore the
Elena L [17]

The wave equation is missing and it is y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

Answer:

A) 0.0534 seconds

B) 0.67N

C) 41

D) (8.50 mm)cos(172 rad/m x + 4830 rad/s t)

Explanation:

we are given weight of string = 0.0125N

Thus, since weight = mg

Then, mass of string = 0.0125/9.8

Mass of string = 1.275 x 10⁻³ kg

Length of string; L= 1.5 m .

mass per unit length; μ = (1.275 x 10⁻³)/1.5

μ = 0.85 x 10⁻³ kg/m

We are given the wave equation: y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

Now if we compare it to the general equation of motion of standing wave on a string which is:

y(x,t) = Acos(Kx − ω t)

We can deduce that

angular velocity;ω = 4830 rad/s

Wave number;k = 172 rad/m

A) Velocity is given by the formula;

V = ω/k

Thus, V = 4830/172 m/s

V = 28.08 m /s

Thus time taken to go up the string = 1.5/28.08 = 0.0534 seconds

B) We know that in strings,

V² = F/μ

Where μ is mass per unit length and V is velocity.

Thus, F = V²*μ =28.08² x 0.85 x 10⁻³

F = 0.67N

C) Formula for wave length is given as; wave length;λ = 2π /k

λ = 2 x π/ 172

λ = 0.0365 m

Thus, number of wave lengths over whole length of string

= 1.5/0.0365 = 41

D) The equation for waves traveling down the string

= (8.50 mm)cos(172 rad/m x + 4830 rad/s t)

8 0
3 years ago
The tendency for an object to remain at rest in continue in motion is called:
nekit [7.7K]

Answer:

A Inertia

Explanation:

7 0
3 years ago
Read 2 more answers
an object has a length of 5.0 cm, aheight of 3.0 cm and a width of 15.0 cm. it has a mass of 24 grams. what is its density?
kondaur [170]
5x3=N
Nx15=F
Fx24=Your Awnser
5 0
2 years ago
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