How many stars are in a galaxy? hundreds thousands millions billions.

Suppose a certain car supplies a constant deceleration of A meter per second per second. If it is traveling at 90km/hr. When. th

aksik [14]

Answer:

i)-6.25m/s

ii)18 metres

iii)26.5 m/s or 95.4 km/hr

Explanation:

Firstly convert 90km/hr to m/s

90 × 1000/3600 = 25m/s

(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)

0 = (25)^2 + 2A(50)

0 = 625 + 100A....then moved the other value to one

-625 = 100A

Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)

(ii) Firstly convert 54km/hr to m/s

In which this is 54 × 1000/3600 = 15m/s

then apply the same formula as that in (i)

0 = (15)^2 + 2(-6.25)s

-225 = -12.5s

Hence the stopping distance = 18metres

(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question

0 = u^2 + 2(-6.25)(56)

u^2 = 700

Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s

In km/hr....26.5 × 3600/1000 = 95.4 km/hr

3 0

2 years ago

A bowling ball that has a radius of 11.0 cm and a mass of 5.00 kg rolls without slipping on a level lane at 2.80 rad/s.

NemiM [27]

Answer:

$\dfrac{K_t}{K_r}=\dfrac{5}{2}$

Explanation:

Given that,

Mass of the bowling ball, m = 5 kg

Radius of the ball, r = 11 cm = 0.11 m

Angular velocity with which the ball rolls, $\omega=2.8\ rad/s$

To find,

The ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball.

Solution,

The translational kinetic energy of the ball is :

$K_t=\dfrac{1}{2}mv^2$

$K_t=\dfrac{1}{2}m(r\omega)^2$

$K_t=\dfrac{1}{2}\times 5\times (0.11\times 2.8)^2$

The rotational kinetic energy of the ball is :

$K_r=\dfrac{1}{2}I \omega^2$

$K_r=\dfrac{1}{2}\times \dfrac{2}{5}mr^2\times \omega^2$

$K_r=\dfrac{1}{2}\times \dfrac{2}{5}\times 5\times (0.11)^2\times (2.8)^2$

Ratio of translational to the rotational kinetic energy as :

$\dfrac{K_t}{K_r}=\dfrac{5}{2}$

So, the ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball is 5:2

4 0

3 years ago

Which ball has the greater average speed during the 1-s interval after release (assuming neither hits the ground during that tim

notka56 [123]

Answer:

The ball thrown downward

Explanation:

When the ball is thrown vertically, the acceleration of it is the gravity acceleration independent if it is thrown downward or upward. However, the acceleration is a vector, so, when the ball is thrown upward, the movement is against the gravity, so the acceleration is negative, and so, the velocity decreases during time; and when the ball is thrown downward, the movement goes to the gravity, so the acceleration is positive, so the velocity increase after time passes.

3 0

2 years ago

Look at your data. With three foods, Flock v

wolverine [178]

Answer:the first one was x

the second one is y

Explanation:

6 0

2 years ago

Urgent!!!!!! 20 points!!!!!!<br> how are the speed, wavelength and frequency of a wave related

zheka24 [161]

Answer:

The higher the frequency, the shorter the wavelength. Because all light waves move through a vacuum at the same speed, the number of wave crests passing by a given point in one second depends on the wavelength. Speed shows how long it takes for wavelengths to travel.

6 0

2 years ago