Garen is climbing stairs to the castle. He travels a distance of 350 stairs in 5 minutes of time.

A 36,287 kg truck has a momentum of 907,175 kg • . What is the truck’s velocity?

Snowcat [4.5K]

By definition,
Momentum = Mass * Velocity

Let v = the velocity of the truck, m/s
The mass of the truck is 36,287 kg.
The momentum is 907,175 (kg-m)/s.

Therefore
907,175 (kg-m)/s = (36287 kg)*(v m/s)
v = 907175/36287 = 25 m/s

Answer: 25 m/s

7 0

3 years ago

Read 2 more answers

A horizontal object-spring system oscillates with an amplitude of 2.8 cm. If the spring constant is 275 N/m and object has a mas

Lisa [10]

Answer:

(a) the mechanical energy of the system, U = 0.1078 J

(b) the maximum speed of the object, Vmax = 0.657 m/s

(c) the maximum acceleration of the object, a_max = 15.4 m/s²

Explanation:

Given;

Amplitude of the spring, A = 2.8 cm = 0.028 m

Spring constant, K = 275 N/m

Mass of object, m = 0.5 kg

(a) the mechanical energy of the system

This is the potential energy of the system, U = ¹/₂KA²

U = ¹/₂ (275)(0.028)²

U = 0.1078 J

(b) the maximum speed of the object

$V_{max} =\omega*A= \sqrt{\frac{K}{M} } *A\\\\V_{max} = \sqrt{\frac{275}{0.5} } *0.028\\\\V_{max} = 0.657 \ m/s$

(c) the maximum acceleration of the object

$a_{max} = \frac{KA}{M} \\\\a_{max} = \frac{275*0.028}{0.5}\\\\a_{max} = 15.4 \ m/s^2$

6 0

3 years ago

An amusement park ride consists of a large vertical cylinder that spins about its axis fast enough that a person inside is stuck

ElenaW [278]

Let N be the normal force that forces the person against the wall.

Then u N = m g is the frictional force supporting the person's weight

and N = m g / u

also, N = m v^2 / R is the normal force providing the centripetal acceleration

So, m g / u = m v^2 / R

v^2 = g R / u

since v = 2 pi R T

4 pi^2 R^2 T^2 = g R / u and T^2 = g / (4 u pi^2 R)

T = 1/ (2 pi) (g /(u R))^1/2 = .159 * (9.8 m/s^2 / (.521 * 4.4 m)) ^1/2

T = .68 / s

Do you see any thing wrong here?

T should have units of seconds not 1 / seconds

v should be 2 * pi * R / T where T is the time for 1 revolution

So you need to make that correction in the above formula for v.

7 0

3 years ago

A horizontal force of 150 N is used to push a 40.0-kg packing crate a distance of 6.00 m on a rough horizontal surface. If the c

icang [17]

Answer:

a. 900 J

b. 0.383

Explanation:

According to the question, the given data is as follows

Horizontal force = 150 N

Packing crate = 40.0 kg

Distance = 6.00 m

Based on the above information

a. The work done by the 150-N force is

$W = F x = \mu N x = \mu\ m\ g\ x$

$W = 150 \times 6$

= 900 J

b. Now the coefficient of kinetic friction between the crate and surface is

$\mu = \frac {F}{m\timesg}$

$= \frac{150}{40\times 9.8}$

= .383

We simply applied the above formulas so that each one part could calculate

3 0

3 years ago

A mountain 10.0 km from a person exerts a gravitational force on him equal to 2.00% of his weight. (a) Calculate the mass of the

LiRa [457]

Answer:

a) M = 2,939 10¹⁷ kg , b) $M_{e}$ / M = 2 10⁷

Explanation:

a) The equation for gravitational force is

F = G m M / r²

Where G is the gravitational constant that is worth 6.67 10⁻¹¹ N m² / kg², m the mass epa person. M the mass of the Mountain and r the distance between them.

The value of this force is 2% of the person's weight

F = 0.02 W = 0.02 mg

we replace

0.02 mg = G m M / r²

M = 0.02 g r² / G

r = 10 km = 10 10³ m = 1.0 10⁴ m

M = 0.02 9.8 (10⁴)² / 6.67 10⁻¹¹

M = 2,939 10¹⁷ kg

b) to compare the masses we find their relationship

$M_{e}$ / M = 5.98 1024 / 2,939 1017

$M_{e}$ / M = 2 10⁷

c) treating the mountain as a point object

d) The mountain is not spherical so the distance changes depending on the height of the mountain

8 0

3 years ago