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faust18 [17]
2 years ago
9

A dog walking to the right at 1.5\,\dfrac{\text m}{\text s}1.5sm​1, point, 5, space, start fraction, m, divided by, s, end fract

ion spies a cat ahead, and begins chasing the cat with a constant acceleration of 12\,\dfrac{\text m}{\text s^2}12s2m​12, space, start fraction, m, divided by, s, start superscript, 2, end superscript, end fraction.
What is the velocity of the dog after running for 3.0\,\text m3.0m3, point, 0, space, m?

​
Physics
1 answer:
Tatiana [17]2 years ago
6 0

Answer:

8.6 m/s

Explanation:

We can find the final velocity of the dog by using the following SUVAT equation:

v^2-u^2=2ad

where

u is the initial velocity

a is the acceleration

d is the distance covered

For the dog in the problem, we have

u = 1.5 m/s

a = 12 m/s^2

And the distance covered is

d = 3.0 m

Therefore, we can re-arrange the equation to find the final velocity, v:

v=\sqrt{u^2+2ad}=\sqrt{1.5^2+2(12)(3.0)}=8.6 m/s

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in physics lab, a cube slides down a frictionless incline as shown in the figure below, and elastically strikes another cube at
Tema [17]
<span>In the physics lab, a cube slides down a frictionless incline as shown in the figure below, check the image for the complete solution:

</span>

3 0
2 years ago
An archer shoots an arrow with a velocity of 45.0m/s at an angle of 50.0degrees with the horizontal.An assistant standing on the
garri49 [273]

Answer:

a) u = 30.29 m/s

b) t = 2.09 s

Explanation:

given,

velocity = 45 m/s

angle (θ) = 50°

horizontal velocity = 45 cos 50°

time taken to reach 150 m.

times = \dfrac{150}{45 cos 50^0}

t  =  5.19 s

a) height of arrow

s = u t +\dfrac{1}{2}gt^2

s = v sin \theta \times t+\dfrac{1}{2}gt^2

s = 45 sin 50^0 \times 5.19 -\dfrac{1}{2}\times 9.81\times 5.19^2

s = 46.78 m

v² - u² = 2 g s

u² = 2 × 9.81 × 46.78

u = 30.29 m/s

b) time taken by the apple = \dfrac{u}{g}=\dfrac{30.29}{9.81}

                                             = 3.09 s

time after which it has to be thrown = 5.19-3.09 = 2.1 s

5 0
2 years 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
2 years ago
Read 2 more answers
Help immediately! Answer I need help.
garri49 [273]

Answer:

0.80 m

Explanation:

elastic potential energy formula

elastic potential energy = 0.5 × spring constant × (extension) 2

4 0
2 years ago
How does 'g' vary from place to place?​
r-ruslan [8.4K]

Explanation:

The acceleration g varies by about 1/2 of 1 percent with position on Earth's surface, from about 9.78 metres per second per second at the Equator to approximately 9.83 metres per second per second at the poles.

8 0
2 years ago
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