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sasho [114]
3 years ago
6

A mechanic pushes a 2.10 â 103-kg car from rest to a speed of v, doing 4,710 j of work in the process. during this time, the car

moves 30.0 m. neglecting friction between car and road, find v and the horizontal force exerted on the car. (a) the speed v m/s (b) the horizontal force exerted on the car (enter the magnitude.) n
Physics
1 answer:
Pachacha [2.7K]3 years ago
3 0
Purches that much and see what your answer is
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) The magnitude of scalar product of two unit vectors perpendicular to each other is ​
fenix001 [56]

This question involves the concept of the scalar product.

The magnitude of the scalar product will be "0".

<h3>SCALAR PRODUCT</h3>

The scalar product, also known as the dot product of the two vectors is given by the following formula:

A.B = |A||B|Cos\theta

where,

  • A.B = Scalar product = ?
  • |A| = Mangnitude of vector A = 1 unit
  • |B| = Magnitude of Vector B = 1 unit
  • θ = Angle between vectors = 90°

Therefore,

A.B = (1)(1)Cos90^o = (1)(1)(0)

A.B = 0

Learn more about scalar product here:

brainly.com/question/6849226

5 0
3 years ago
A 500 lines per mm diffraction grating is illuminated by light of wavelength 580 nm . what is the maximum diffraction order seen
vampirchik [111]

The maximum diffraction order seen is 3.

<h3>What is the maximum diffraction order seen?</h3>

We know that the maximum angle of diffraction Q_m of the furthest bright fringe < Q = 90 degrees.

Here we need to compute the nth bright fringe for which is approximated to 90 degrees.

The angle of nth bright fringe is given by;

sin(Q_m) = n(λ)N

Approximating Q_m ≈ 90 degrees.

sin (90) = nλN

n = sin (90) / (λN)

n = 1 / ((580 x 10⁻⁶)500)

n = 3.5 orders

Since, we knew that Q_m < 90 degrees, we will choose n = 3 as the maximum number of orders.

Thus, the maximum diffraction order seen is 3.

Learn more about maximum diffraction here: brainly.com/question/14703089

#SPJ4

6 0
2 years ago
A ball rolls downhill with a constant acceleration of 4m/s squared. If it started from rest,it’s velocity at the end of 3 meters
vladimir2022 [97]

Answer:

4.9 m/s

Explanation:

Since the motion of the ball is a uniformly accelerated motion (constant acceleration), we can solve the problem by using the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For the ball in this problem,

u = 0 (it starts from rest)

a=4 m/s^2 is the acceleration

s = 3 m is the distance covered

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(4)(3)}=4.9 m/s

3 0
4 years ago
a conical pendulum is formed by attaching a 50g mass to a 1.2m string. The mass swings around the circle of radius 25cm (a) calc
kipiarov [429]
A. 0.5 miles an hour 
B. 0.2 miles an hour
4 0
3 years ago
Determine the gravitational field 300km above the surface of the earth. How does this compare to the field on the earth's surfac
Serjik [45]
The strength of the gravitational field is given by:
g= \frac{GM}{r^2}
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.

In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
r=R+h=6370 km+300 km=6670 km= 6.67 \cdot 10^{6} m

And now we can use the previous equation to calculate the field strength at that altitude:
g= \frac{GM}{r^2}= \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(6.67 \cdot 10^6 m)^2}  = 8.95 m/s^2

And we can see this value is a bit less than the gravitational strength at the surface, which is g_s = 9.81 m/s^2.
4 0
3 years ago
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