All categories

3 years ago

A 700 kg car makes a turn going at 30 m/s with radius of

Physics

1 answer:

Gnoma [55]3 years ago

5 0

Answer:5250 N

Explanation: ig:iihoop.vince

You might be interested in

You're driving down the highway late one night at 22 m/s when a deer steps onto the road 38 m in front of you. Your reaction tim

sleet_krkn [62]

speed of the car is given on the road

$v = 22 m/s$

reaction time is given as

$t = 0.50 s$

now we can find the distance that it will move during this reaction time

$d_1 = 22* 0.50 = 11 m$

now the deceleration of the car is 10 m/s^2

so the distance that it will move before stop is given by

$v_f^2 - v_i^2 = 2 a d$

$0^2 - 22^2 = 2*(-10)*d$

$d = 24.2 m$

so the total distance that it require to stop is given as

$d = 24.2 + 11 = 35.2 m$

while the deer is standing at distance 38 m

so the car will stop 2.8 m before the position of deer

3 0

3 years ago

Can someone help me please

makvit [3.9K]

Answer:

Time, I believe. Pretty sure it's time lol

3 0

3 years ago

Martha was leaning out of the window on the second floor of her house and speaking to Steve. Suddenly, her glasses slipped from

DiKsa [7]

Answer:

s = 23.72 m

v = 21.56 m/s²

Explanation:

given

time to reach the ground (t) = 2.2 second

we know that

a) s = u t + 0.5 g t²

u = 0 m/s

g = 9.8 m/s²

s = 0 + 0.5 × 9.8 × 2.2²

s = 23.72 m

b) impact velocity

v = √(2gh)

v = √(2× 9.8 × 23.72)

v = √464.912

v = 21.56 m/s²

6 0

3 years ago

Please help with both questions thank you!

Katen [24]

I think is:

(4)Inner core

(4)different Kingdome different species

hope this help :)

3 0

2 years ago

Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated p

Marta_Voda [28]

Answer:

Acceleration, $a=9.91\times 10^{15}\ m/s^2$

Explanation:

It is given that,

Separation between the protons, $r=3.73\ nm=3.73\times 10^{-9}\ m$

Charge on protons, $q=1.6\times 10^{-19}\ C$

Mass of protons, $m=1.67\times 10^{-27}\ kg$

We need to find the acceleration of two isolated protons. It can be calculated by equating electric force between protons and force due to motion as :

$ma=\dfrac{kq^2}{r^2}$

$a=\dfrac{kq^2}{mr^2}$

$a=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{1.67\times 10^{-27}\times (3.73\times 10^{-9})^2}$

$a=9.91\times 10^{15}\ m/s^2$

So, the acceleration of two isolated protons is $9.91\times 10^{15}\ m/s^2$ . Hence, this is the required solution.

3 0

3 years ago