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Usimov [2.4K]
1 year ago
6

calculate the deceleration of a bullet initially travelling at 360 m/s, if it is brought to rest after travelling 8.5 cm into a

tree trunk which it has struck.
Physics
1 answer:
sp2606 [1]1 year ago
8 0

The de - acceleration of bullet from 360 m/s after it struck the tree trunk and was brought to rest is -152.54 x 10⁴ m/s².

<h3>What is Acceleration in Kinematics?</h3>

Acceleration is defined as the rate of change of velocity with respect to time. Mathematically -

<h3>a = Δv/Δt</h3>

Given is a bullet initially travelling at 360 m/s struck into a tree trunk 8.5 cm inside and comes to rest. From this, we can write

initial velocity [u] = 360 m/s

final velocity [v] = 0 m/s

distance travelled [d] = 8.5 cm = 0.085 m

The time span after which bullet comes to rest will be -

Δt = d/u

Δt = 0.085/360 = 2.36 x 10 ⁻⁴ seconds

a = Δv/Δt

Δv = 0 - 360 = - 360 m/s

a = (v - u)/Δt

a = - 360/(2.36 x 10 ⁻⁴)

a = - 152.54 x 10⁴ m/s²

Therefore, the de - acceleration of bullet from 360 m/s after it struck the tree trunk and was brought to rest is -152.54 x 10⁴ m/s².

To solve more questions on Kinematics, visit the link below-

brainly.com/question/14413005

#SPJ1

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Two men decide to use their cars to pull a truck stuck in the mud. They attach ropes and one pulls with a force of 615 N at an a
stiv31 [10]

Answer:

1398.12 N

Explanation:

We define the x-axis in the direction parallel to the movement of the truck  on and the y-axis in the direction perpendicular to it.

x-components  of the ropes forces

T₁x = 615N*cos31°=527.1579 N  :Tension in direction x of the rope of the car 1

T₂x= 961 N*cos25°=870.96 N  :Tension in direction x of the rope of the car 2

Net forward force exerted on the truck in the direction it is headed (Fnx)

Fnx = T₁x  + T₂x

Fnx = 527.1579 N  + 870.96 N

Fnx = 1398.12 N

4 0
4 years ago
A 2530-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that
gavmur [86]

Answer:

A = 1.4 m/s²

B = -0.10493 m/s³

a = 1.29507 m/s²

T = 28095.8271 N

T = 1.13198 W

Explanation:

t = Time taken

g = Acceleration due to gravity = 9.81 m/s²

The equation

v(t)=At+Bt^2

Differentiating with respect to time

\frac{dv}{dt}=\frac{d(At+Bt^2)}{dt}\\\Rightarrow 1.4=A+2Bt

At t = 0

1.4=A

Hence, A = 1.4 m/s²

B=\frac{v-At}{t^2}\\\Rightarrow B=\frac{2.18-1.4\times 1.8}{1.8^2}\\\Rightarrow B=-0.10493\ m/s^3

B = -0.10493 m/s³

At t = 5 seconds

a=1.4+2\times -0.010493\times 5=1.29507\ m/s^2

a = 1.29507 m/s²

T=m(a+g)\\\Rightarrow T=2530(1.29507+9.81)\\\Rightarrow T=28095.8271\ N

T = 28095.8271 N

Weight of rocket

W=2530\times 9.81=24819.9\ N

\frac{T}{W}=\frac{28095.8271}{24819.9}\\\Rightarrow \frac{T}{W}=1.13198\\\Rightarrow T=1.13198W

T = 1.13198 W

3 0
3 years ago
A cyclotron is used to produce a beam of high-energy deuterons that then collide with a target to produce radioactive isotopes f
serg [7]

Answer:

19080667.0818 m/s

0.637294 m

2.1875\times 10^{15}

Explanation:

m = Mass of deuterons = 3.34\times 10^{-27}\ kg

v = Velocity

K = Kinetic energy = 3.8 MeV

d = Diameter

B = Magnetic field = 1.25 T

q = Charge of electron = 1.6\times 10^{-19}\ C

t = Time = 1 s

i = Current = 350 μA

Kinetic energy is given by

K=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{\dfrac{2K}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 3.8\times 10^6\times 1.6\times 10^{-19}}{3.34\times 10^{-27}}}\\\Rightarrow v=19080667.0818\ m/s

The speed of the deuterons when they exit is 19080667.0818 m/s

In this system the centripetal and magnetic force will balance each other

\dfrac{mv^2}{r}=qvB\\\Rightarrow \dfrac{mv^2}{\dfrac{d}{2}}=qvB\\\Rightarrow d=\dfrac{2mv}{qB}\\\Rightarrow d=\dfrac{2\times 3.34\times 10^{-27}\times 19080667.0818}{1.6\times 10^{-19}\times 1.25}\\\Rightarrow d=0.637294\ m

The diameter is 0.637294 m

Current is given by

i=\dfrac{nq}{t}\\\Rightarrow n=\dfrac{it}{q}\\\Rightarrow n=\dfrac{350\times 10^{-6}\times 1}{1.6\times 10^{-19}}\\\Rightarrow n=2.1875\times 10^{15}

The number of deuterons is 2.1875\times 10^{15}

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3 years ago
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e-lub [12.9K]

Answer:

sorry I had find answer from everywhere but can't find

Explanation:

can u can send it with some information please

4 0
2 years ago
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Alborosie

Answer:

Strato-volcano

Explanation:

Strato-volcanoes are usually characterized by the presence of steep-sided slopes, with distinct craters, and are frequently erupted and conical in appearance. This type of volcano is generally felsic in nature. Due to the presence of high silica content, the magma being highly viscous, moves at a relatively slower rate. These are highly explosive and produce a large number of pyroclastic materials, lava flow, volcanic ashes, and gases.

They are also commonly considered as the composite volcano, and are comprised of alternating tephra and solidified lava layers.

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3 years ago
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