Can you help quick in science please

Rupert (66 kg) is now in a full-pipe of radius 25 m, calculate the minimum speed at which Rupert can skate to ensure he will not

V125BC [204]

at the top most point if Rupert will not fall then normal force at the top point is almost zero for minimum speed

so here we can say

$F_n + mg = m\frac{v^2}{R}$

now if

$F_n = 0$

$mg = \frac{mv^2}{R}$

$v = \sqrt{Rg}$

$v = \sqrt{25 \times 9.8}$

$v = 15.65 m/s$

so above will be the minimum speed

4 0

3 years ago

Please tell me the order the answers go. Best and most correct answer gets Brainliest.

iogann1982 [59]

See the attached picture:

8 0

3 years ago

Question 8 (1 point) How many protons are in an element with an atomic number of 8 and a mass number of 18? a 3 b 8 c 10 d 18

yarga [219]

Answer: B - 8

Explanation: 8 protons because number of protons is equal to number of atoms in the nucleus.

6 0

3 years ago

The student soon loses his balance and falls backwards off the board at a velocity of 1.0 m/s. Assuming momentum is conserved in

Phantasy [73]

Answer:

v2 = 27.3m/s

Explanation:

Assuming forward as positive.

Mass = m1 = 64kg

Let v be the common velocity of the student and the skateboard.

mass of skateboard = m2 = 5.94kg

v = 1.4m/s

Since the skateboard and the student are initially moving together at the same velocity their momentum together is

(m1 + m2)v

Let the final velocity of the student be v1 and the final velocity of the skateboard be v2

v1 = – 1.0m/s (falls backwards that's why the velocity is negative since we are assuming forward as positive)

Then from conservation of momentum, momentum before is equal to momentum after.

(m1 + m2)v = m1v1 + m2v2

m2v2= (m1 + m2)v – m1v1

v2 = ( (m1 + m2)v – m1v1)/m2

v2 = ( (64 + 5.94)×1.4 – 64×(-1.0))/5.94

v2 = ( (64 + 5.94)×1.4 + 64×1.0)/5.94

v2 = 27.3m/s

5 0

3 years ago

A certain light truck can go around a flat curve having a radius of 150 m with a maximum speed of 35.5 m/s. a) What is the coeff

postnew [5]

Answer:

The coefficient of friction present between the roadway and the wheels of the truck is <u>0.833</u>.

Explanation:

Given:

Radius of the curve (R) = 150 m

Maximum speed of truck (v) = 35.5 m/s

Let the coefficient of friction between the roadway and the wheels of the truck be "μ".

As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.

The centripetal force acting on the truck is given as:

$F_c=\frac{mv^2}{R}$

Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.

Frictional force is given as:

$f=\mu N$