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Likurg_2 [28]
3 years ago
8

A bullet with a mass of 0.020 kg collides with a 2.5 kg wooden block at

Physics
1 answer:
kondaur [170]3 years ago
5 0

Answer:

151.2 m/s

Explanation:

m1 = 0.020 kg

m2 = 2.5 kg

vf = 1.2 m/s

m1v1 +m2v2 = (m1 +m2)vf

0.020v1 + 0 = (0.020 +2.5)(1.2)

---> 151.2 m/s

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Identify the velocity-versus-time plot(s) that correspond to motion under a constant, non-zero acceleration.
ICE Princess25 [194]

Answer:

See explanation

Explanation:

The question is incomplete because the images were not attached but I will try to help you as much as possible.

Constant acceleration implies that the velocity increases uniformly with time.

The graph of constant acceleration is a straight line graph having a slope. The slope of the graph is constant at any point along the straight line.

The image attached shows a velocity-time graph depicting constant acceleration.

5 0
3 years ago
The ash produced when solid waste is incinerated is ______ than the original waste.
11111nata11111 [884]
The ash is more toxic
5 0
3 years ago
Which object is moving faster?<br> A
Amiraneli [1.4K]

Answer:

A

Explanation:

4 0
3 years ago
Read 2 more answers
Hola tengo un taller de física y nose como resolver este pregunta
Blizzard [7]

a) 0.26 h

b) 71.4 km

Explanation:

a)

In order to solve the problem, we have to know what is the final velocity of the car.

Here, we assume that the final velocity reached by the car is

v=300 km/h

Therefore, we can find the time taken by the car to reach this velocity by using the suvat equation:

v=u+at

where:

u = 250 km/h is the initial velocity

a=190 km/h^2 is the acceleration of the car

v = 300 km/h is the final velocity

t is the time

Solving for t, we find:

t=\frac{v-u}{a}=\frac{300-250}{190}=0.26 h

b)

In order to find the distance covered by the car, we can use the following suvat equation:

s=ut+\frac{1}{2}at^2

where:

s is the distance covered

u is the initial velocity

a is the acceleration

t is the time

For the car in this problem, we have:

u = 250 km/h

t = 0.26 h (calculated in part a)

a=190 km/h^2

Therefore, the distance covered is

s=(250)(0.26)+\frac{1}{2}(190)(0.26)^2=71.4 km

8 0
3 years ago
A 6-in-wide polyamide F-1 flat belt is used to connect a 2-in-diameter pulley to drive a larger pulley with an angular velocity
Likurg_2 [28]

Answer:

a) Fc = 4.15 N, Fi = 435.65 N, (F1)a = 640 N, and F2  = 239.6 N,

b) Ha = 1863.75 N, nfs = 1 , length = 11.8 mm

Explanation:

Given that:

γ= 9.5 kN/m³ = 9500N/m3

b = 6 inches = 0.1524 m

t = 0.0013 mm

d = 2 inches  = 0.0508 m

n = 1750 rpm

H_{nom}=2hp=1491.4W

L = 9 ft = 2.7432 m

Ks = 1.25

g = 9.81 m/s²

a)

w=\gamma b t = 9500* 0.1524*0.0013=1.88N/m

V=\frac{\pi d n}{60} =\pi *0.0508*1750/60=4.65 m/s

F_c=\frac{wV^2}{g}=1.88*4.65^2/9.81=4.15N

(F_1)_a=bF_aC_pC_v=0.1524*6000*0.7*1=640N

T=\frac{H_{nom}n_dK_s}{2\pi n}= \frac{1491*1.25*1}{2*\pi*1750/60}=10.17Nm

F_2=(F_1)_a-\frac{2T}{D}= 640-\frac{2*10.17}{0.0508} =239.6N

F_i=\frac{(F_1)_a+F_2}{2} -F_c=435.65N

b)

H_a=1491*1.25=1863.75W

n_f_s=\frac{H_a}{H_{nom}K_S }=1

dip = \frac{L^2w}{8F_i} =\frac{2.7432*1.88}{435.65}=11.8mm

7 0
3 years ago
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