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1 year ago

A cat accelerates from rest to 10m/s when it sees a dog. This takes 2 seconds. What was the acceleration of the cat

Physics

2 answers:

Nookie1986 [14]1 year ago

5 0

The formula is a= chance in velocity/time
A=10-0/2
A=10/2
A=5 m/s^2 (meters per second squared)

nlexa [21]1 year ago

4 0

Answer :

Acceleration of the cat is $a=5\ m/s^2$

Explanation:

It is given that,

Initial speed of the cat, u =0 (at rest)

Final speed of the cat, v = 10

Time taken, t = 2 s

Let a is the acceleration of the car. We know that the rate of change of velocity of an object is called its acceleration. Mathematically, it is given by :

$a=\dfrac{v-u}{t}$

$a=\dfrac{10-0}{2}$

$a=5\ m/s^2$

So, the acceleration of the cat was $5\ m/s^2$ . Hence, this is the required solution.

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Light with a wavelength of 495 nm is falling on a surface and electrons with a maximum kinetic energy of 0.5 eV are ejected. Wha

devlian [24]

Answer:

To increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.

Explanation:

First, we have to calculate the work function of the element. The maximum kinetic energy as a function of the wavelength is given by:

$K_{max}=\frac{hc}{\lambda}-W$

Here h is the Planck's constant, c is the speed of light, $\lambda$ is the wavelength of the light and W the work function of the element:

$W=\frac{hc}{\lambda}-K_{max}\\W=\frac{(4.14*10^{-15}eV\cdot s)(3*10^8\frac{m}{s})}{495*10^{-9}m}-0.5eV\\W=2.01eV$

Now, we calculate the wavelength for the new maximum kinetic energy:

$W+K_{max}=\frac{hc}{\lambda}\\\lambda=\frac{hc}{W+K_{max}}\\\lambda=\frac{(4.14*10^{-15}eV\cdot s)(3*10^8\frac{m}{s})}{2.01eV+1.5eV}\\\lambda=3.54*10^{-7}m=354*10^{-9}m=354nm$

This wavelength corresponds to ultraviolet radiation. So, to increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.

8 0

2 years ago

A community calendar allows nonprofit organizations to add events and their intended purposes for community members to see.

lbvjy [14]

Out of all given choices, activities and donor list can be found on the given kind of informative website.

Answer: Option A and B

<u>Explanation:</u>

Non-profit need to do the examination and for sake activities that simply aren't successful. And afterwards, they have to look to a portion of these auxiliary changes that to discover increasingly productive approaches to make a feasible money related model for their social change work.

When occasions are crucial, allowed to visit, and concentrated on developing as well as managing present or potential significant donors (people, establishments, corporate pioneers, they can bode well. Yet, only in the event that you catch up with participants on a one-on-one premise to additionally put them in the association and in the long run request that they contribute or reestablish their commitments.

On the off chance, you don't charge those to visit with the goal that can approach them for a greater, and progressively important blessing not far off.

8 0

1 year ago

This is a problem about a child pushing a stack of two blocks along a horizontal floor. The masses of the blocks, and the coeffi

Rufina [12.5K]

Answer:

N = 23.4 N

Explanation:

After reading that long sentence, let's solve the question

The contact force is the so-called normal in this case we can find it by writing the translational equilibrium equation for the y axis

N - w₁ -w₂ =

N = m₁ g + m₂ g

N = g (m₁ + m₂)

let's calculate

N = 9.8 (0.760 + 1.630)

N = 23.4 N

This is the force of the support of the two blocks on the surface.

7 0

1 year ago

A force F of strength 20 N acts on an object of mass 3 kg as it moves a distance of 4 m. If F is perpendicular to the 4 m displa

forsale [732]

Work = (displacement) x (force in the direction of the displacement)

Since none of the force is in the direction of the displacement, the work it does
is zero. Something else is making the object move. It's not the 20N force.

3 0

1 year ago

Read 2 more answers

In this chapter, we treated force as a push or pull, now we say it is part of an interaction. Is a force a push or pull, or part

mafiozo [28]

Answer:

A force can be described as both.

A force is a push or pull upon an item coming about because of the object's connection with another object.

8 0

1 year ago