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3 years ago

A cat is moving at 18 m/s when it accelerates for 2 seconds. What is its acceleration?

Physics

1 answer:

vova2212 [387]3 years ago

5 0

Answer: acceleration: $4 m/s^{2}$

velocity: $26 m/s$

Explanation:

The complete question is written bellow:

A cat is moving at 18 m/s when it accelerates at $4 m/s^{2}$ for 2 seconds. What is his new velocity?

In this situation the following equation will be useful:

$V=V_{o}+at$

Where:

$V$ is the cat’s final velocity (new velocity)

$V_{o}=18 m/s$ is the cat’s initial velocity

$a=4 m/s^{2}$ is the cat's acceleration

$t=2 s$ is the time

Solving the equation:

$V=18 m/s+(4 m/s^{2})(2 s)$

$V=26 m/s$ This is the cat's new velocity

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One object is at rest, and another is moving. The two collide in a one-dimensional, completely inelastic collision. In other wor

Bess [88]

Answer:

a) The final velocity is 20 m/s when the large-mass object is the one moving initially.

b) The final velocity is 9.0 m/s when the small-mass object is the one moving initially.

Explanation:

The momentum of the system is calculated as the sum of the momenta of each object. Each momentum is calculated as follows:

p = m · v

Where:

p = momentum.

m = mass.

v = velocity.

Then, the momentum of the system is the following:

m1 · v1 + m2 · v2 = (m1 + m2) · v

Where:

m1 = mass of the bigger object.

v1 = velocity of the bigger object.

m2 = mass of the smaller object.

v2 = velocity of the smaller object.

v = final velocity of the two objects after the collision.

Solving the equation for the final velocity:

(m1 · v1 + m2 · v2)/ (m1 + m2) = v

a) Let´s calculate the final velocity when the bigger object is moving:

(7.1 kg · 29 m/s + 3.2 kg · 0)/(7.1 kg + 3.2 kg) = v

v = 20 m/s

b) When the smaller object is moving:

(7.1 kg · 0 m/s + 3.2 kg · 29 m/s) / (7.1 kg + 3.2 kg) = v

v = 9.0 m/s

7 0

3 years ago

What is (9x10^9)(2.6x10^-6)(1.4x10^-6) / 36

Natali [406]

Answer:

0.00091

Explanation:

(9x10^9) (2.6x10^-6) (1.4x10^-6) / 36

(9,000,000,000) (0.0000026) (0.0000014) /36

23,400(0.0000014) /36

0.03276 /36

0.00091

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3 years ago

6 A test of a driver's perception/reaction time is being conducted on a special testing track with level, wet pavement and a dri

mylen [45]

Answer:

a. 10.5 s b. 6.6 s

Explanation:

a. The driver's perception/reaction time before drinking.

To find the driver's perception time before drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)

a = - 499.52 m²/s²/234.7 m

a = -2.13 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (0 m/s - 22.35 m/s)/-2.13 m/s²

t = - 22.35 m/s/-2.13 m/s²

t = 10.5 s

b. The driver's perception/reaction time after drinking.

To find the driver's perception time after drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)

a = 179.83 m²/s² - 499.52 m²/s²/234.7 m

a = -319.69 m²/s² ÷ 234.7 m

a = -1.36 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²

t = - 8.94 m/s/-1.36 m/s²

t = 6.6 s

4 0

3 years ago

A physics student tests the theory of projectile motion by leaping off a 225 meter tall building. She runs off the building hori

tamaranim1 [39]

In this item, we are given with the x-component of the velocity. The y-component is equal to 0 m/s. The time it takes for it to reach the volume can be related through the equation,

d = V₀t + 0.5gt²

Substituting the known values,

225 = (0 m/s)(t) + (0.5)(9.8)(t²)

Simplifying,

t = 6.776 s

To determine the distance of the student from the edge of the building, we multiply the x-component by the calculated time.

range = (12.5 m/s)(6.776 s)

range = 84.7 m

Answer: 84.7 m

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3 years ago

Glass is transparent to visibile light under normal conditions; however, at extremely high intensities, glass will absorb most o

8_murik_8 [283]

Answer:

3 photons

Explanation:

The energy of a photon E can be calculated using this formula:

$E=\frac{hc}{\lambda}$

Where $h$ corresponds to Plank constant (6.626070x10^-34Js), $c$ is the speed of light in the vacuum (299792458m/s) and $\lambda$ is the wavelength of the photon(in this case 800nm).