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Yuki888 [10]
3 years ago
8

Sledding down a hill, you are traveling at 10 m/s when you reach the bottom. You (inertia 80 kg ) then move across horizontal sn

ow toward a 400-kg boulder but jump off the sled (inertia 8.0 kg ) the instant before it hits the boulder. The boulder is sitting on very slick ice and moves freely when the sled hits it. The sled bounces back, moving at 6.0 m/s. At what speed does the boulder move after the sled hits it?
Physics
1 answer:
RSB [31]3 years ago
5 0

Answer:

V = 0.32 m /s

Explanation:

Momentum of the man on the sledge along with sledge

= (80+8) x 10 = 880 kg. m/s When the man jumps off the sledge , the velocity of the sledge will remain intact at 10 m/s

When the sledge hits the boulder and bounces back the momentum of the sledge - boulder system will remain conserved .

change in momentum of sledge = 8 x (10 +6 )

= 128  kg m/s

Change in the momentum of boulder

= 400 V -0

= 400V

400V = 128

V = 0.32 m /s

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Answer:

F = 24 N

Explanation:

In this exercise we have a bar l = 100 m with a center of gravity x = 4 m, which force is needed to lift it from the other end

Let's use the rotational equilibrium relationship, where we consider the counterclockwise rotations as positive and fix the reference system at the point closest to the center of gravity

           ∑ τ = 0

           F l -x W = 0

           F = \frac{x}{l} \  W

let's calculate

          F = \frac{4}{100}4/100 600

          F = 24 N

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A block lies on a horizontal frictionless surface. A horizontal force of 100 N is applied to the block giving rise to an acceler
KIM [24]

Answer:

(a) m = 33.3 kg

(b) d = 150 m

(c) vf = 30 m/s

Explanation:

Newton's second law to the block:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

Data

F= 100 N

a= 3.0 m/s²

(a) Calculating of the  mass of the block:

We replace dta in the formula (1)

F = m*a

100 =  m*3

m = 100 / 3

m = 33.3 kg

Kinematic analysis

Because the block  moves with uniformly accelerated movement we apply the following formulas:

d= v₀t+ (1/2)*a*t² Formula (2)

vf= v₀+a*t   Formula (3)

Where:  

d:displacement in meters (m)  

t : time interval in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

Data

a= 3.0 m/s²

v₀= 0

t = 10 s

(b) Distance the block will travel if the force is applied for 10 s

We replace dta in the formula (2):

d= v₀t+ (1/2)*a*t²

d = 0+ (1/2)*(3)*(10)²

d =150 m

(c) Calculate the speed of the block after the force has been applied for 10 s

We replace dta in the formula (3):

vf= v₀+a*t

vf= 0+(3*(10)

vf= 30 m/s

4 0
3 years ago
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