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3 years ago

A motorboat traveling on a straight course slows

Physics

1 answer:

never [62]3 years ago

3 0

Answer:

2.572 m/s²

Explanation:

Convert the given initial velocity and final velocity rates to m/s:

65 km/h → 18.0556 m/s
35 km/h → 9.72222 m/s

The motorboat's displacement is 45 m during this time.

We are trying to find the acceleration of the boat.

We have the variables v₀, v, a, and Δx. Find the constant acceleration equation that contains all four of these variables.

v² = v₀² + 2aΔx

Substitute the known values into the equation.

(9.72222)² = (18.0556)² + 2a(45)
94.52156173 = 326.0046914 + 90a
-231.4831296 = 90a
a = -2.572

The magnitude of the boat's acceleration is |-2.572| = 2.572 m/s².

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A 5000-kg freight car crashes into a 10,000-kg freight car at rest. They couple upon collision and move with a speed of 2 m/s. W

nirvana33 [79]

Answer:

6 m/s

Explanation:

mass of moving car m1=5000 kg

initial velocity of moving car vi1=?

mass of car at rest = m2=10000 kg

initial velocity of car at rest = vi2=0

final velcoities of both cars after collision = vf1=vf2= 2m/s

using conservation of momentum rule

m1vi1+m2vi2=m1vf1+m2vf2

putting values

==> 5000 × vi1 + 1000 × 0 = 5000 × 2 + 10000 × 2

==> 5000 ×vi1 = 2 × 15000

==> vi1 = 2 × 15000 ÷ 5000

==> vi1= 2×3=6 m/s

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3 years ago

A train that travel 100 kilometer in 4hours is traveling at what speed?

DochEvi [55]

Speed = Distance/Time = 100 km / 4 hours = 100/4 km per hour = 25 kph

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3 years ago

What does K stand for in Hookes law?

wlad13 [49]

So Hooke's law says that that law is proportional to how much I stretch the spring. Alright. So f=kx<span>. x is the length of the spring now minus its length when it's relaxed and nobody's pulling on it. k is a constant called the spring constant.</span>

3 0

3 years ago

Read 2 more answers

If Star A is magnitude 1.0 and Star B is magnitude 9.6 , which is brighter and by what factor?

ludmilkaskok [199]

Answer:

Star A is brighter than Star B by a factor of 2754.22

Explanation:

Lets assume,

the magnitude of star A = m₁ = 1

the magnitude of star B = m₂ = 9.6

the apparent brightness of star A and star B are b₁ and b₂ respectively

Then, relation between the difference of magnitudes and apparent brightness of two stars are related as give below: $(m_{2} - m_{1}) = 2.5\log_{10}(b_{1}/b_{2})$

The current magnitude scale followed was formalized by Sir Norman Pogson in 1856. On this scale a magnitude 1 star is 2.512 times brighter than magnitude 2 star. A magnitude 2 star is 2.512 time brighter than a magnitude 3 star. That means a magnitude 1 star is (2.512x2.512) brighter than magnitude 3 bright star.

We need to find the factor by which star A is brighter than star B. Using the equation given above,

$(9.6 - 1) = 2.5\log_{10}(b_{1}/b_{2})$

$\frac{8.6}{2.5} = \log_{10}(b_{1}/b_{2})$

$\log_{10}(b_{1}/b_{2}) = 3.44$

Thus,

$(b_{1}/b_{2}) = 2754.22$

It means star A is 2754.22 time brighter than Star B.

3 0

3 years ago

A Van de Graaff generator is one of the original particle accelerators and can be used to accelerate charged particles like prot

Rzqust [24]

Answer:

595391.482946 m/s

$3.21875\times 10^{6}$

Explanation:

E = Energy = 1.85 keV

I = Current = 5.15 mA

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

t = Time taken = 1 second

m = Mass of proton = $1.67\times 10^{-27}\ kg$

Velocity of proton is given by

$v=\sqrt{\dfrac{2E}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 1.85\times 10^3\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}}\\\Rightarrow v=595391.482946\ m/s$

The speed of the proton is 595391.482946 m/s

Current is given by

$I=\dfrac{\Delta Q}{t}\\\Rightarrow \Delta Q=It\\\Rightarrow \Delta Q=5.15\times 10^{-3}\times (1\ sec)\\\Rightarrow Q=5.15\times 10^{-3}\ C$

Number of protons is

$n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{5.15\times 10^{-3}}{1.6\times 10^{-19}}\\\Rightarrow n=3.21875\times 10^{6}\ protons$

The number of protons is $3.21875\times 10^{6}$

3 0

4 years ago