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2 years 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]2 years 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]2 years 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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The current in a long solenoid of radius 6 cm and 17 turns/cm is varied with time at a rate of 5 A/s. A circular loop of wire of

jonny [76]

Answer:

53.63 μA

Explanation:

radius of solenoid, r = 6 cm

Area of solenoid = 3.14 x 6 x 6 = 113.04 cm^2 = 0.0113 m^2

n = 17 turns / cm = 1700 /m

di / dt = 5 A/s

The magnetic field due to the solenoid is given by

B = μ0 n i

dB / dt = μ0 n di / dt

The rate of change in magnetic flux linked with the solenoid =

Area of coil x dB/dt

= 3.14 x 8 x 8 x 10^-4 x μ0 n di / dt

= 3.14 x 64 x 10^-4 x 4 x 3.14 x 10^-7 x 1700 x 5 = 2.145 x 10^-4

The induced emf is given by the rate of change in magnetic flux linked with the coil.

e = 2.145 x 10^-4 V

i = e / R = 2.145 x 10^-4 / 4 = 5.36 x 10^-5 A = 53.63 μA

6 0

1 year ago

A pitcher delivers a fast ball with a velocity of 43 m/s to the south. the batter hits the ball and gives it a velocity of 51 m/

hoa [83]

Assuming north as positive direction, the initial and final velocities of the ball are:
$v_i=-43 m/s$ (with negative sign since it is due south)
$v_f=+51 m/s$
the time taken is $t=1.0 ms=0.001 s$ , so the average acceleration of the ball is given by
$a= \frac{v_f-v_i}{t}= \frac{51 m/s-(-43 m/s)}{0.001 s}=9.4 \cdot 10^4 m/s^2$
And the positive sign tells us the direction of the acceleration is north.

4 0

1 year ago

The 600-N ball shown is suspended on a string AB and rests against the frictionless vertical wall. The string makes an angle of

Harrizon [31]

Answer: T = 692.82 and 346.4 N

Explanation:

Given that;

w = 600 N

∅ = 30°

ΣFy = ma

a = 0 m/s²

ΣF = T(cos30°) - W = 0

T(cos30°) = W

we Divide both sides by cos30°

T = W / cos30o

T= 600N / cos30°

T = 692.82

and ∑fx

F = T sin∅

F = 692.82 × (sin30°)

F = 346.4 N

4 0

2 years ago

If isomerization requires breaking the pi bond, what minimum energy is required for isomerization in j/mol?

aliina [53]

<span>The minimum energy required for isomerization is 267 000 J/mol
</span>

The isomerization of cis-but-2-ene to trans-but-2-ene requires breaking of the π bond.

The bond energy of a C-C σ bond is 347 kJ/mol.

The bond energy of a C=C double bond (σ + π) is 614 kJ/mol.

So the bond energy of a π bond is (614 – 347) kJ/mol = 267 kJ/mol =
267 000 J/mol.

6 0

1 year ago

A 43.9-g piece of copper (CCu= 0.385 J/g°C) at 135.0°C is plunged into 254 g of water at 39.0°C. Assuming that no heat is lost t

Semmy [17]

Answer:

$T = 40.501\,^{\textdegree}C$

Explanation:

The interaction of the piece of copper and water means that the first one need to transfer heat in order to reach a thermal equilibrium with water. Then:

$-Q_{out,Cu} = Q_{in,H_{2}O}$

After a quick substitution, the expanded expression is:

$-(43.9\,g)\cdot (0.385\,\frac{J}{g\cdot ^{\textdegree}C} )\cdot (T-135^{\textdegree}C) = (254\,g)\cdot (4.187\,\frac{J}{g\cdot ^{\textdegree}C} )\cdot (T-39\,^{\textdegree}C)$

$-16.902\,\frac{J}{^{\textdegree}C}\cdot (T-135^{\textdegree}C) = 1063.498\,\frac{J}{^{\textdegree}C} \cdot (T-39^{\textdegree}C)$

$43758,192\,J = 1080.4\,\frac{J}{^{\textdegree}C}\cdot T$

The final temperature of the system is:

$T = 40.501\,^{\textdegree}C$

8 0

1 year ago

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