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ira [324]
2 years ago
15

Two vectors are presented as a=3.0i +5.0j and b=2.0i+4.0j find (a) a x b, ab (c) (a+b)b and (d) the component of a along the dir

ection of
b.
Physics
1 answer:
Svet_ta [14]2 years ago
3 0
Let's ask this question step by step:
 Part A) 
 a x b = (3.0i + 5.0j) x (2.0i + 4.0j) = (12-10) k = 2k
 ab = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
 Part (c)
 (a + b) b = [(3.0i + 5.0j) + (2.0i + 4.0j)]. (2.0i + 4.0j)
 (a + b) b = (5.0i + 9.0j). (2.0i + 4.0j)
 (a + b) b = 10 + 36
 (a + b) b = 46
 Part (d)
 comp (ba) = (a.b) / lbl
 a.b = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
 lbl = root ((2.0) ^ 2 + (4.0) ^ 2) = root (20)
 comp (ba) = 26 / root (20)
 answer
 2k
 26
 46
 26 / root (20)
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A kinesin that is transporting a secretory vesicle uses approximately 80 ATP molecules/s. Each ATP provides a kinesin molecule w
Firdavs [7]

Answer:

The force is  F  =   8*10^{-12} \ N

Explanation:

From the question we are told that

     The rate at which ATP molecules are used is R =  80 ATP/ s

       The energy provided by a single ATP is  E_{ATP} =  0.8 *  10^{-19} J

       The velocity of the kinesin is  v  =  800 nm/s =  800*10^{-9} m/s

The power provided by the ATP in one second is  mathematically represented as

       P =  E_{ATP}  *  R

substituting values

       P =  80 * 0.8*10^{-19 }

       P =  6.4 *10^{-18}J/s

Now  this power is mathematically represented as

       P  =  F *  v

Where  F  is  the force the kinesin is exerting

  Thus  

          F  =   \frac{P}{v}

substituting values

            F  =   \frac{6.4*0^{-18}}{800 *10^{-9}}

           F  =   8*10^{-12} \ N

7 0
2 years ago
A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,00
mote1985 [20]

Answer:

1. 6.99x 10^-6V/m

2. 18m

Explanation:

See attached file

7 0
2 years ago
: The truck is to be towed using two ropes. Determine the magnitudes of forces FA and FB acting on each rope in order to develop
Sholpan [36]

Answer:

Fa=774 N

Fb=346 N

Explanation:

We will solve this problem by equating forces on each axis.

  1. On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
  2. On y-axis let forces in positive y-direction be positive and forces in negative y-direction be negative

While towing we know that car is mot moving in y-direction so net force in y-axis must be zero

⇒∑Fy=0

⇒Fa*sin(50)-Fb*sin(20)=0

⇒Fa*sin(50)=Fb*sin(20)

⇒Fa=2.24Fb

Given that resultant force on car is 950N in positive x-direction

⇒∑Fx=950  

⇒Fa*cos(20)+Fb*cos(50)=950

⇒2.24*Fb*cos(20)+Fb(50)=950

⇒Fb*(2.24*cos(20)+cos(50))=950

⇒Fb=\frac{950}{2.24*cos(20)+cos(50)}

⇒Fb=\frac{950}{2.24*0.94+0.64}

⇒ Fb=\frac{950}{2.75}=345.5

⇒Fa=2.24*Fb

      =2.24*345.5

      =773.93

Therefore approximately, Fa=774 N and Fb=346 N

5 0
2 years ago
Two objects have masses m and 5m, respectively. They both are placed side by side on a frictionless inclined plane and allowed t
poizon [28]

Answer:

(E) The two objects reach the bottom of the incline at the same time.

Explanation:

Given;

first object with mass, m

second object with mass, 5m

The acceleration of gravity for both object is the same = 9.8 m/s²

Since both objects have the same acceleration of gravity, and no external force due friction (frictionless inclined plane), they will reach bottom of the inclined at the time.

Thus, the acceleration due to gravity is constant for all objects regardless of their masses.

Therefore, the correct option is E;

(E) The two objects reach the bottom of the incline at the same time.

5 0
2 years ago
The magnitude J(r) of the current density in a certain cylindrical wire is given as a function of radial distance from the cente
kipiarov [429]

Answer:

I=68.31\times 10^{-6}\ A

Explanation:

Given that

J(r) = Br

We know that area of small element

dA = 2 π dr

I = J A

dI = J dA

Now by putting the values

dI = B r . 2 π dr

dI= 2π Br² dr

Now by integrating above equation

\int_{0}^{I}dI= \int_{r_1}^{r_2}2\pi Br^2 dr

I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}

Given that

B= 2.35 x 10⁵ A/m³

r₁ = 2 mm

r₂ = 2+ 0.0115 mm

r₂ = 2.0115 mm

I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}

By putting the values

I={2\pi \times 2.35 \times 10^5 }\times \dfrac{(2.0115\times 10^{-3})^3-(2\times 10^{-3})^3}{3}\ A

I=68.31\times 10^{-6}\ A

7 0
2 years ago
Read 2 more answers
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