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Basile [38]
3 years ago
13

Which of the following is the dimensional formula of the quantity which is expressed as force per unit area? *

Physics
1 answer:
MAVERICK [17]3 years ago
5 0

Answer:

[MLT⁻²], the first option, is the answer.

Explanation:

Pressure is the unit expressed as force per unit area.

The dimensional formula of pressure is [MLT⁻²].

Thus the answer would be option 1!

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On a Saturday afternoon, you decide to pay a neighborhood kid to mow your lawn. The kid usesa manual push lawn mower with a mass
Mars2501 [29]

Answer with Explanation:

We are given that

A.Mass,m=12 kg

\theta=53^{\circ}

\mu_k=0.16

Speed,v=1.5m/s

Net force in x direction must be zero

F_{net}=0

Fsin\theta-f=0

Fsin\theta=f

Net force in y direction

N-mg-Fcos\theta=0

N=mg+Fcos\theta

f=\mu_kN=\mu_k(mg+Fcos\theta)

Fsin\theta=\mu_k(mg+Fcos\theta)

Fsin\theta=\mu_kmg+\mu_kFcos\theta

Fsin\theta-\mu_kFcos\theta=\mu_kmg

F(sin\theta-\mu_kcos\theta)=\mu_kmg

F=\frac{\mu_kmg}{sin\theta-\mu_kcos\theta}

Power,P=Fv

P=\frac{\mu_kmg}{sin\theta-\mu_kcos\theta}v

Where g=9.8m/s^2

B.Substitute the values

P=\frac{0.16\times 12\times 9.8}{sin53-0.16cos53}\times 1.5

P=40.17W

6 0
4 years ago
Now we’ll use the component method to add two vectors. We will use this technique extensively when we begin to consider how forc
dimaraw [331]

Answer:

The magnitude of the vector sum of A and B is 65.8 cm and its direction 61.6°

Explanation:

Since vector A has magnitude 50 cm and a direction of 30, its x - component is A' = 50cos30 = 43.3 cm and its y - component is A" = 50sin30 = 25.

Also, Since vector B has magnitude 35 cm and a direction of 110, its x - component is A' = 35cos110 = -11.97 cm and its y - component is A" = 35sin110 = 32.89 cm.

So, the vector sum R = A + B

The x-component of the vector sum is R' = A'+ B' = 43.3 cm + (-11.97 cm) = 43.3 cm - 11.97 cm = 31.33 cm

The y-component of the vector sum is R" = A"+ B" = 25 cm + 32.89 cm = 57.89 cm

So, the magnitude of R = √(R'² + R"²)

= √((31.33 cm)² + (57.89 cm)²)

= √(981.5689 cm² + 3,351.2521 cm²)

= √(4,332.821 cm²)

= 65.82 cm

≅ 65.8 cm

The direction of R is Ф = tan⁻¹(R"/R')

= tan⁻¹(57.89 cm/31.33 cm)

= tan⁻¹(1.84775)

= 61.58°

≅ 61.6°

So, the magnitude of the vector sum of A and B is 65.8 cm and its direction 61.6°

4 0
3 years ago
A truck using a rope to tow a 2230-kg car accelerates from rest to 13.0 m/s in a time of 15.0s. How strong must the rope be? μk
Leokris [45]

Answer:

The rope must have a force of 10084,21 N

Explanation

Acceleration calculation

The car acceleration is equal to the acceleration of the truck

ac: car acceleration\frac{m}{s^{2} }

at: truck acceleration\frac{m}{s^{2} })

ac = at= \frac{vf-vi}{t-ti}  equation(1)

Known information:

vi = Initial speed = 0, ti = initial time = 0

vf = Final speed = 13 \frac{m}{s}, t = final time =5 s

We replaced the known information in the equation(1):

ac = at = \frac{13-0}{15-0}

ac=ac=\frac{13}{15}  \frac{m}{s}

Dynamic analysis

The forces acting on the car are the following:

Wc: Car weight

N: normal force, road force on the car

Ff: Friction force

T: Force of tension

Car weight calculation:

Wc=mc*g

mc = Car mass = 2230kg

g = Gravity acceleration=9.8 \frac{m}{s^{2} }

Wc= 2230*9.8

Wc=21854 N

Normal force calculation:

Newton's first law

sum Fy= 0

N-W=0

N=W

N=21854 N

Friction force calculation (Ff):

We have the formula to calculate the friction force:

Ff = μk * N  Equation (3)

μk kinetic coefficient of friction

We know that μk = 0.373and N= 21854N ,then:

Ff=0.373*21854

Ff=8151.54 N

Calculation of the tension force in the rope (T):

Newton's Second law

sum Fx= mc*ac

T-Ff=mc*ac

T=2230(\frac{13}{15}) + 8151.54

T=10084,21 N

Answer: The rope must have a force of 10084,21 N

8 0
3 years ago
The temperature of a gas is _____.
kykrilka [37]
Directly proportional to pressure
8 0
3 years ago
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Is 250 000 miles from the earth to the moon" is a qualitative<br> Observation<br> TRUE<br> Or false
lys-0071 [83]

Answer:

True

Explanation:

3 0
3 years ago
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