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3 years ago

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A cat chase a mouse across a 1.0 m high table. The mouse steps out of the way, and the car slides off the table and strikes the

floor 2.2m from the edge of the table. When the car slid off the table what was it’s speed?
Answer is 4.9 m/s how do we get 4.9 m/s?

1 answer:

Tema [17]3 years ago

5 0

The cat fell 1.0 m from the ground.

Using the formula

$h = v_{oy}t - \frac{gt^2}{2}$

Here, $v_{oy} = 0, \; h = 1.0 \; m.$

Solving for t, the time it spent in the air is

$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2(1.0 \; m)}{9.8 \; m/s^2}} = 0.451753951 \; s$

The cat does not accelerate along the horizontal, so it has constant horizontal velocity. Since it strikes the floor 2.2 m from the table, then

$v_x = \Delta x/t = \frac{2.2\; m}{0.451753951 s} = 4.869907597 \; m/s \Rightarrow 4.9 \; m/s$

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At 19 ºC a volume of air takes us 9.0 mL in an expandable container. What is its volume after it is heated up to 50. ºC?

Katena32 [7]

According to Charles law, we know, at constant pressure, volume is directly proportional to temperature.
So, <span>V/T = constant
</span>
V₁/t₁ = V₂/t₂
V₁t₂ = V₂t₁

Here, we have: V₁ = 9 mL
V₂ = ?
T₂ = 50+272 = 323 K
T₁ = 19+273 = 292 K

Substitute their values into the expression:
9 × 323 = V₂ × 292
V₂ = 2907 / 292
V₂ = 9.95
After rounding-off to unit place value, it would be equal to 10 mL

So, In short Option C would be your correct answer.

Hope this helps!

5 0

3 years ago

A roller coaster loaded with passengers has a mass of 2000 kg; the radius of curvature of the track at the lowest point of the t

Tom [10]

Answer:

The force is exerted on the vehicle by the track is 46600 N

Explanation:

It is given that,

Mass of the roller coaster, m = 2000 kg

Radius of the circular track, r = 24 m

Tangential speed of the vehicle, v = 18 m/s

Let W is the weight of the roller coaster, F is the force is exerted on the vehicle by the track and $F_c$ is the centripetal force. So,

$F-mg=\dfrac{mv^2}{r}$

$F=m(\dfrac{v^2}{r}+g)$

$F=2000\times (\dfrac{(18)^2}{24}+9.8)$

F = 46600 N

So, the net force exerted on the vehicle by the track is 46600 N. Hence, this is the required solution.

7 0

3 years ago

Ball thrown upwards at velocity of 21.3 m/s what time will ball reach highest point

Nutka1998 [239]

Recall that

${v_y}^2-{v_{0y}}^2=2a_y(y-y_0)$

At its highest point, the ball will have 0 vertical velocity, so that

$-\left(21.3\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-9.80\,\dfrac{\mathrm m}{\mathrm s^2}\right)y_{\mathrm{max}}$

$\implies y_{\mathrm{max}}=23.1\,\mathrm m$

4 0

3 years ago

A fireworks shell is accelerated from rest to a velocity of 56.0 m/s over a distance of 0.270 m.

deff fn [24]

Answer:

(a) The acceleration lasts 9.642 seconds.

(b) The acceleration of the fireworks shell is 5.808 meters per square second.

Explanation:

Statement is incomplete. Complete description is presented below:

<em>A fireworks shell is accelerated from rest to a velocity of 56 meters per second over a distance of 0.270 meters. </em><em>(a)</em><em> How long (in seconds) did the acceleration last? </em><em>(b)</em><em> Calculate the acceleration.</em>

(b) Let suppose that the fireworks shell is accelerated uniformly. Given that initial ( $v_{o}$ ) and final speed ( $v_{f}$ ), measured in meters per second, and distance ( $s$ ), measured in meters, are known, we calculate the acceleration ( $a$ ), measured in meters per square second, by this kinematic expression:

$a = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot s}$ (1)

If we know that $v_{o} = 0\,\frac{m}{s}$ , $v_{f} = 56\,\frac{m}{s}$ and $s = 0.270\,m$ , then the acceleration is:

$a = \frac{\left(56\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot (270\,m)}$

$a = 5.808\,\frac{m}{s^{2}}$

The acceleration of the fireworks shell is 5.808 meters per square second.

(a) Now we calculate the time associated with acceleration ( $t$ ), measured in seconds, by means of this kinematic equation:

$t = \frac{v_{f}-v_{o}}{a}$

If we know that $v_{o} = 0\,\frac{m}{s}$ , $v_{f} = 56\,\frac{m}{s}$ and $a = 5.808\,\frac{m}{s^{2}}$ , then time associated with acceleration is:

$t = \frac{56\,\frac{m}{s}-0\,\frac{m}{s}}{5.808\,\frac{m}{s^{2}} }$

$t = 9.642\,s$

The acceleration lasts 9.642 seconds.

8 0

3 years ago

monochromatic light of wavelength 500 nm is incident normally on a diffraction grating. if the third order maximum is 32. how ma

Alona [7]

Answer:

The total number of maxima that can be seen is 11

Explanation:

Given the data in the question

wavelength λ = 500 nm = 5 × 10⁻⁷ m

if the third order maximum is 32

i.e m = 3 and θ = 32°

Now, we know that condition for diffraction maximum is as follows;

d × sinθ = m × λ

so we substitute in our given values

d × sin( 32° ) = 3 × 5 × 10⁻⁷ m

d × sin( 32° ) = 1.5 × 10⁻⁶ m

d = [ 1.5 × 10⁻⁶ m ] / sin( 32° )

d = 2.83 × 10⁻⁶ m

Now, maxima n when θ = 90° will be;

sin( 90° ) = nλ / d

1 = nλ / d

d = nλ

n = d / λ

we substitute

n = [ 2.83 × 10⁻⁶ m ] / [ 5 × 10⁻⁷ m ]

n = 5.66

so 5 is the max value

hence, total maxima value is;

⇒ 2n + 1 = 2( 5 ) + 1 = 10 + 1 = 11

Therefore, total number of maxima that can be seen is 11

5 0

3 years ago