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2 years ago

What is the acceleration of a ball rolling down a ramp that starts from rest and travels 0.9 m in 3 s?

1 answer:

3 0

Given:
u = 0, initial velocity
s 0.9 m, distance traveled.
t = 3 s, the time taken.

Let a = the acceleration. Then
s = ut + (1/2)*a*t²
(0.9 m) = 0.5*(a m/s²)*(3 s)²
0.9 = 4.5a
a = 0.2 m/s²

Answer: 0.2 m/s²

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If you wished to warm 100 kg of water by 15 degrees celsius for your bath, how much heat would be required? (give your answer in

Anit [1.1K]

For the answer to the question above,
Q = amount of heat (kJ) 
cp = specific heat capacity (kJ/kg.K) = 4.187 kJ/kgK 
m = mass (kg) 
dT = temperature difference between hot and cold side (K). Note: dt in °C = dt in Kelvin 

Q = 100kg * (4.187 kJ/kgK) * 15 K 
Q = 6,280.5 KJ = 6,280,500 J = 1,501,075.5 cal

6 0

2 years ago

What is the y component of a vector defined as 12.2m at 81.5°?

sergejj [24]

Answer:

Explanation:

This is a displacement vector since it is defined in terms of distance (meters, to be exact). The way you find the y-component is

$V_y=Vsin\theta$ which says that you multiply the magnitude of the vector (its length) by the sin of the direction (the angle):

$V_y=12.2sin(81.5)$ and get

$V_y=$ 12.1 m

3 0

2 years ago

An archer fires an arrow, which produces a muffled "thwok" as it hits a target. If the archer hears the "thwok" exactly 1 s afte

aniked [119]

Answer:

35,79 meters

Explanation:

So, we got an archer, and we got a target. Lets call the distance between this two d.

Now, the archer fires the arrow, that, in a time $t_{arrow}$ travels the distance d with a speed $v_{arrow}$ of 40 m/s and hits the target. We can see that the equation will be:

$v_{arrow} * t_{arrow} = d\\ \\40 \frac{m}{s} * t_{arrow} = d$

Immediately after this, the arrow produces a muffled sound, which will travel the distance d at 340 m/s in a time $t_{sound}$ . Obtaining :

$v_{sound} * t_{sound} = d\\ \\340 \frac{m}{s} * t_{sound} = d$ .

Finally, the sound reaches the archer, exactly 1 second after he fired the bow, so:

$t_{arrow} + t _{sound} = 1 s$ .

This equation allows us to write:

$t _{sound} = 1 s - t_{arrow}$ .

Plugging this relationship in the distance equation for the sound:

$340 \frac{m}{s} * t_{sound} = d \\ \\ 340 \frac{m}{s} * (1 s- t_{arrow}) = d$ .

Now, we can replace d from the first equation, and obtain:

$40 \frac{m}{s} * t_{arrow} = d \\ 40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * (1 s- t_{arrow})$ .

Now, we can just work a little bit:

$40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * 1 s - 340 \frac{m}{s} * t_{arrow} \\ \\ 40 \frac{m}{s} * t_{arrow} + 340 \frac{m}{s} * t_{arrow} = 340 m \\ \\ 380 \frac{m}{s} * t_{arrow} = 340 m \\ \\ t_{arrow} = \frac{340 m}{380 \frac{m}{s}} \\ \\ t_{arrow} = 0.8947 s$ .

Now, we can just plug this value into the first equation:

$40 \frac{m}{s} * t_{arrow} = d$

$40 \frac{m}{s} * 340/380 s = 35,79 s = d$

6 0

2 years ago

A 10-meter long ramp has a mechanical advantage of 5. What is the height of the ramp?

denpristay [2]

1. If the ramp has a length of 10 and has a mechanical advantage (MA) of 5. Then we need to find the height of the ramp.
Formula:
MA = L / H
Since we already have the mechanical advantage and length, this time we need to find the height .
MA 5 = 10 / h
h = 10 / 5
h = 2 meters

Therefore, the ramp has a length of 10 meters, a height of 2 meters with a mechanical advantage of 5.

6 0

2 years ago

A boy uses a slingshot to launch a pebble straight up into the air. The pebble reaches a height of 37.0 m above the launch point

denis-greek [22]

Answer:

V0=27.4 m/s; t=0.8 s

Explanation:

Final position y=37.0 m, time = 2.3 s; Initial position is set to be zero. We calculate the initial speed with the kinematics equation:

$y_f=v_0t-0.5*g*t^2$ We solve for initial speed