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Sloan [31]
2 years ago
15

How large a force is required to accelerate a 1300 kg car from rest to a speed of 20 m/s in a distance of 80 m?

Engineering
2 answers:
topjm [15]2 years ago
7 0

F=m*a

F=80*20

F =1600 ans"

Stolb23 [73]2 years ago
5 0

Answer: basically we are asked to find it's force and the formula that helps us to find force is Force= Mass× Acceleration

so let's put out the given and what we need to find

GIVEN. Solution

F=m.a

Mass= 1300. F=1300kg×20m/s

speed (Acceleration)=20m/2. F=26000Kg.m/s

Force=? F=26000N

N stands for Newton and the reason that it comes Newton is because it is the SI-unit of Force and the result of kg×m/s.

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Explain how use of EGR is effective in reducing NOx emissions 4. In most locations throughout the U.S., the octane number of reg
TiliK225 [7]

Answer:please see attached file

Explanation:

3 0
3 years ago
The error in the trapezoidal rule is proportional to h^2 True b. False
Lynna [10]

Answer:

b) False

Explanation:

in trapezoidal rule the error is proportional to h^3 and the order of accuracy is proportional to h^3 .

Trapezoidal rule is numerical integration method .Trapezoidal rule is used to find the area of curves.In trapezoidal rule we finds the approximate value of integration.But the real value of integration will not differ to much from the value which finds by using trapezoidal rule.

4 0
3 years ago
If gain of the critically damped system is increased, the system will behave as a) Under damped b) Over damped c) Critically dam
Ganezh [65]

Answer:

a) Under damped

Explanation:

Given that system is critically damped .And we have to find out the condition when gain is increased.

As we know that damping ratio given as follows

\zeta =\dfrac{C}{C_c}

Where C is the damping coefficient and Cc is the critical damping coefficient.

C_c=2\sqrt{mK}

So from above we can say that

\zeta =\dfrac{C}{2\sqrt{mK}}

\zeta \alpha \dfrac{1}{\sqrt K}

From above relationship we can say when gain (K) is increases then system will become under damped system.

7 0
3 years ago
A single Gates Rubber V belt is to be selected to deliver engine power to the wheel-drive transmission of a riding tractor. A 5-
beks73 [17]

Answer:

Hello there, please check step by step explanations to get answers.

Explanation:

Given that:

5-hp single-cylinder engine is used. At most, the belt transmits 60 percent of this power. The driving sheave has a diameter of 6.2 in. and the driven, 12.0 in. The belt selected should be as close to 92 in pitch length as possible. The engine speed is governor-controlled to a maximum of 3100 rev/min. Select a satisfactory belt, and specify it using the standard designation.

See attached documents for clearity and step by step procedure to answer

7 0
3 years ago
Block A hangs by a cord from spring balance D and is submerged in a liquid C contained in beaker B. The mass of the beaker is 1.
nikitadnepr [17]

Answer:

a)  m_e= 3.05 Kg

b)  \rho=1072.3kg/m^3

c)  m_e= 3.05 Kg

Explanation:

From the question we are told that:

Beaker Mass m_b=1.20

Liquid Mass m_l=1.85

Balance D:

Mass m_d=3.10

Balance E:

Mass m_e=7.50

Volume v=4.15*10^{-3}m^3

a)

Generally the equation for Liquid's density is mathematically given by

m_e=m_b+m_l+(\rho*v)

\rho=\frac{7.50-(1.2+1.85)}{4.15*10^{-3}}

\rho=1072.3kg/m^3

b)

Generally the equation for D's Reading at A pulled is mathematically given by

m_d = mass of block - mass of liquid displaced

m_d=m- (\rho *v )

m=3.10+ (1072.30 *4.15*10^{-3}m^3 )

m=18.10kg

c)

Generally the equation for E's Reading at A pulled is mathematically given by

m_e=m_b+m_l

m_e = 1.20 + 1.85

m_e= 3.05 Kg

6 0
3 years ago
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