A 35 grams bullet travels with a velocity of magnitude 126 km/h. What is the bullet's linear momentum?

A 1000-kg car is moving at 30 m/s around a horizontal unbanked curve whose diameter is 0.20 km. What is the magnitude of the fri

omeli [17]

Answer:

4500 N

Explanation:

When a body is moving in a circular motion it will feel an acceleration directed towards the center of the circle, this acceleration is:

a = v^2/r

where v is the velocity of the body and r is the radius of the circumference:

Therefore, a body with mass m, will feel a force f:

f = m v^2/r

Therefore we need another force to keep the body(car) from sliding, this will be given by friction, remember that friction force is given a the normal times a constant of friction mu, that is:

fs = μN = μmg

The car will not slide if f = fs, i.e.

fs = μmg = m v^2/r

That is, the magnitude of the friction force must be (at least) equal to the force due to the centripetal acceleration

fs = (1000 kg) * (30m/s)^2 / (200 m) = 4500 N

7 0

3 years ago

Read 2 more answers

(???) Is iz allergic to peanut butter jigglys, I JIGGLE IN THE WIND!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Vadim26 [7]

Uh are u okay? G tell ur parents i-

7 0

3 years ago

A series of optical telescopes produced an image that has a resolution of about 0.00350 arc second.

Mila [183]

Answer:

The smallest diameter is $D =122 \ m$

Explanation:

From the question we are told that

The resolution of the telescope is $\theta = 0.00350 \ arc \ second$

The wavelength is $\lambda = 1.70 \mu m = 1.70 *10^{-6} \ m$

From the question we are told that

$1 arc \ sec = \frac{1}{3600^o}$

So $0.00350 \ arc \ second = x$

Therefore

$x = 0.00350 * \frac{1}{3600 }$

$x = ( 9.722*10^{-7} )^o$

Now $1^o = \frac{\pi}{180}$

So $(9.722*10^{-7})^o = \theta$

=> $\theta = (9.722*10^{-7}) * \frac{\pi}{180}$

$\theta = 1.69*10^{-8} rad$

The smallest diameter is mathematically represented as

$D = \frac{1.22 \lambda }{\theta }$

substituting values

$D = \frac{1.22 * 1.7 *10^{-6}} {1.69 *10^{-8} }$

$D =122 \ m$

6 0

3 years ago

What is the density of an object with mass of 60 g and a volume of 2cm3

Artemon [7]

Answer:

D=mass /volume

=60/0.002

=30000

4 0

2 years ago

g An airplane is flying through a thundercloud at a height of 1500 m. (This is a very dangerous thing to do because of updrafts,

Stolb23 [73]

Answer:

$523269.9\ \text{N/m}$

Explanation:

q = Charge

r = Distance

$q_1=25\ \text{C}$

$r_1=3000\ \text{m}$

$q_2=40\ \text{C}$

$r_2=850\ \text{m}$

The electric field is given by

$E=E_1+E_1\\\Rightarrow E=k(\dfrac{q_1}{r_1^2}+\dfrac{q_2}{r_2^2})\\\Rightarrow E=9\times 10^9\times (\dfrac{25}{3000^2}+\dfrac{40}{850^2})\\\Rightarrow E=523269.9\ \text{N/m}$

The electric field at the aircraft is $523269.9\ \text{N/m}$

4 0

3 years ago