All categories

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.

You might be interested in

Tom and his little sister are enjoying an afternoon at the ice rink. they playfully place their hands together and push against

Westkost [7]

Newton's third law says:
"For every action, there is an equal and opposite reaction. ".

So, the force that Tom does on the sister is equal to force the sister applies on Tom:
 $F_t = F_s$
where the label "t" means "on Tom", while the label "s" means "on the sister".

From Newton's second law, we also know
 $F=ma$
where m is the mass and a the acceleration. so we can rewrite the first equation as
 $m_t a_t = m_s a_s$
And find Tom's acceleration:
 $a_t = \frac{m_s}{m_t} a_s = \frac{15 kg}{61 kg} (2.1 m/s^2) =0.52 m/s^2$

5 0

2 years ago

A child of mass m is at the edge of a merry-go-round of diameter d. When the merry-go-round is rotating with angular acceleratio

dem82 [27]

Answer:

The torque on the child is now the same, τ.

Explanation:

It can be showed that the external torque applied by a net force on a rigid body, is equal to the product of the moment of inertia of the body with respect to the axis of rotation, times the angular acceleration.
In this case, as the movement of the child doesn't create an external torque, the torque must remain the same.
The moment of inertia is the sum of the moment of inertia of the merry-go-round (the same that for a solid disk) plus the product of the mass of the child times the square of the distance to the center.
When the child is standing at the edge of the merry-go-round, the moment of inertia is as follows:

$I_{to} = I_{d} + m*r^{2} = m*\frac{r^{2}}{2} + m*r^{2} = \frac{3}{2}* m*r^{2} (1)$

So, τ = 3/2*m*r²*α (2)

When the child moves to a position half way between the center and the edge of the merry-go-round, the moment of inertia of the child decreases, as the distance to the center is less than before, as follows:

$I_{t} = I_{d} + m*\frac{r^{2}}{4} = m*\frac{r^{2}}{2} + m*\frac{r^{2}}{4} = \frac{3}{4}* m*r^{2} (3)$

Since the angular acceleration increases from α to 2*α, we can write the torque expression as follows:

τ = 3/4*m*r² * (2α) = 3/2*m*r²

same result than in (2), so the torque remains the same.

7 0

1 year ago

The electric potential in a region that is within 2.00 mm of the origin of a rectangular coordinate system is given by V=Axl+Bym

Fynjy0 [20]

Answer:

Given the potential, $V = Ax^l+By^m+Cz^n+D$

The components of the electric field are:

$E_x = \frac{-dV}{dx} = -Alx^l^-^1$

$E_y = \frac{-dV}{dy} = - Bmy^m^-^1$

$E_z = \frac{-dV}{dz} = - nCzn^n^-^1$

Let's calculate the potential difference for all given points.

$V(0, 0, 0) = 10V => Ax^l+By^m+Cz^n+D = 10$

$=> D = 10$

$V(1, 0, 0) = 4V => A + 10 = 4$

Solving for A, we have:

$A = 4 - 10$

$A = -6$

$V(0, 1, 0) = 6V => B + 10 = 6$

Solving for B, we have:

$B = 6 - 10$

$B = -4$

$V(0, 0, 1) = 8V => C + 10 = 4$

Solving for C, we have:

$C = 8 - 10$

$C = -2$

For all given points, let's calculate the magnitude of electric field as follow:

$E_x (1, 0, 0) = 16 => - Alx^l^-^1 = 16$

$Al = -16$

Solving for l, we have:

$l = \frac{-16}{A}$

From above, A = -6

$l = \frac{-16}{-6}$

$l = \frac{8}{3}$

$E_y (0, 1, 0) = 16=> Bmy^m^-^1 = 16$

$Bm = -16$

Solving for m, we have:

$m = \frac{-16}{A}$

From above, B = -4

$m = \frac{-16}{-4}$

$m = 4$

$E_y (0, 0, 1) = 16=> nCz^n^-^1 = 16$

$nC = - 16$

Solving for n, we have:

$n = \frac{-16}{C}$

From above, C = -2

$n = \frac{-16}{-2}$

$n = 8$

4 0

1 year ago

The blade of metal cutter is shorter than that scissors of tailors

Naily [24]

Explanation:it is beause they are sharper and also have less surface area and therefore more pressure

8 0

2 years ago

a student drew the following model: volcano cooling crust motion plates tension which landform should the student put next in th

Andrei [34K]

Answer:

C) rift valley

Explanation:

A rift valley is a lowland region formed by the interaction of Earth's tectonic plates. This small rift valley has a typical formation—long, narrow, and deep. It was formed by the Thingvellir rift, where the North American and Eurasian tectonic plates are tearing, or rifting, apart over a hotspot on the island of Iceland.

3 0

1 year ago

Read 2 more answers