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2 years ago

What force is required to accelerate a 1840 kg car from 4.77 m/s to 23.5 m/s,

1 answer:

7 0

A :-) for this question , we should apply
a = v - u by t
Given - u = 4.77 m/s
v = 23.5 m/s
t = 5.18 m/s
Solution -
a = v - u by t
a = 23.5 - 4.77
a = 28.27 m/s^2

.:. The acceleration is 28.27 m/s^2

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A trunk is dragged 3.0 meters across an attic floor by a force of 2.0N how much positive work is done on the trunk

Novay_Z [31]

The formula for work is

F*d

Therefore work=2.0N*3.0=6N*m

8 0

3 years ago

What is the mechanical advantage of the level shown below?

snow_lady [41]

the correct answer is B. 1.27

Mechanical advantage of a lever is simply the ratio of the effort arm to the load arm.Effort arm is the distance from the pivot to the point of application of force while load arm is the distance of the lord from the pivot.

therefore, in this question, the effort arm is 0.28m while the load arm is 0.22 m. MA is calculated as follows: MA=effort arm/load arm

=0.28m/0.22m=1.27

5 0

3 years ago

A string that is under 54.0 N of tension has linear density 5.20 g/m . A sinusoidal wave with amplitude 2.50 cm and wavelength 1

kicyunya [14]

Answer:

8.89288275 m/s

Explanation:

F = Tension = 54 N

$\mu$ = Linear density of string = 5.2 g/m

A = Amplitude = 2.5 cm

Wave velocity is given by

$v=\sqrt{\frac{F}{\mu}}\\\Rightarrow v=\sqrt{\frac{54}{5.2\times 10^{-3}}}\\\Rightarrow v=101.90493\ m/s$

Frequency is given by

$f=\frac{v}{\lambda}\\\Rightarrow f=\frac{101.90493}{1.8}\\\Rightarrow f=56.61385\ Hz$

Angular frequency is given by

$\omega=2\pi f\\\Rightarrow \omega=2\pi 56.61385\\\Rightarrow \omega=355.71531\ rad/s$

Maximum velocity of a particle is given by

$v_m=A\omega\\\Rightarrow v_m=0.025\times 355.71531\\\Rightarrow v_m=8.89288275\ m/s$

The maximum velocity of a particle on the string is 8.89288275 m/s

5 0

2 years ago

An airplane flying parallel to the ground undergoes two consecutive displacements. The first is 50 km 60.0° west of north, and t

Rashid [163]

Answer:

LOL

Explanation:

IMAGINE POSTING UR CLASSWORK LOLL

8 0

2 years ago

A small airplane with a wingspan of 18.0 m is flying due north at a speed of 63.6 m/s over a region where the vertical component

choli [55]

Answer:

(a) ε = 1373.8.

(b) The wingtip which is at higher potential.

Explanation:

(a) Finding the potential difference between the airplane wingtips.

Given the parameters

wingspan of the plane is = 18.0m

speed of the plane in north direction is = 70.0m/s

magnetic field of the earth is = 1.20μT

The potential difference is given as:

ε = Blv

where ε = potential difference of wingtips

B = magnetic field of earth

l = wingspan of airplane

v = speed of airplane

ε = 1.2 x 18.0 x 63.6

ε = 1373.8

(b) Which wingtip is at higher potential?

The wingtip which is at higher potential.

5 0

3 years ago