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AfilCa [17]
3 years ago
6

A student pulls a 2.0-kg object to the left with a force of 30 N, while another student is pulling against the object in the opp

osite direction with a force of 20 N. What is the acceleration of the object?
5 m/s2, left

5 m/s2, right

20 m/s2, left

20 m/s2, right
Physics
1 answer:
alex41 [277]3 years ago
5 0

Answer:

5 m/s2, left

Explanation:

We can solve the problem by applying Newton's second law of motion, which  states that:

\sum F=ma

where:

\sum F is the net force acting on an object

m is the mass of the object

a is its acceleration

In this problem, we have:

\sum F=30 N - 20 N = 10 N (to the left) is the net force on the object

m = 2.0 kg is the mass

So, the acceleration is:

a=\frac{\sum F}{m}=\frac{10}{2.0}=5.0 m/s^2

in the same direction as the force (left).

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A ball of mass m on a string of length L is attached to a pivot. The ball is released from rest while the string is parallel to
mash [69]

Answer:

L/2

Explanation:

Neglect any air or other resistant, for the ball can wrap its string around the bar, it must rotate a full circle around the bar. This means the ball should be able to swing to the top position where it's directly above the bar. By the law of energy conservation, this happens when the ball is at the same level as where it's previously released vertically. It means the swinging radius around the bar must be at least half of the string length.

So the distance d between the bar and the pivot should be at least L/2

8 0
3 years ago
5. Dry ice is an example of _________, which is the process of a solid turning directly into a gas. (1 point) precipitation melt
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6 0
3 years ago
If a bus travel 200 km in 45 minutes calculate the speed in kilometre per minute​
erma4kov [3.2K]

Answer:

multiply that and divided by 45

6 0
3 years ago
Read 2 more answers
A 2.5 m -long wire carries a current of 8.0 A and is immersed within a uniform magnetic field B⃗ . When this wire lies along the
leva [86]

Answer:

Explanation:

Let the magnetic field be B = B₁i + B₂j + B₃k

Force = I ( L x B )  , I is current , L is length and B is magnetic field .

In the first case

force = - 2.3 j N

L = 2.5 i

puting the values in the equation above

- 2.3 j = 8 [ 2.5 i x ( B₁i + B₂j + B₃k )]

= - 20 B₃ j + 20 B₂ k

comparing LHS and RHS ,

20B₃ = 2.3

B₃ = .115

B₂ = 0

In the second case

L = 2.5 j

Force = I ( L x B )

2.3i−5.6k = 8 ( 2.5 j x (B₁i + B₂j + B₃k )

=  - 20 B₁ k + 20B₃ i

2.3i−5.6k = - 20 B₁ k + 20B₃ i

B₃ = .115

B₁ = .28

So magnetic field B = .28 i + .115 B₃

Part A

x component of B = .28 T

Part B

y component of B = 0

Part C

z component of B = .115 T .

8 0
3 years ago
Light of wavelength 600 nm passes though two slits separated by 0.22mm and is observed on a screen 1.1m behind the slits. The lo
Ne4ueva [31]

Answer:

y(m=1)=\frac{(1)(600*10^{-9}m)(1.1m)}{0.22m}=3*10^{-6}m\\\\y(m=1)=\frac{(-1)(600*10^{-9}m)(1.1m)}{0.22m}=-3*10^{-6}m

Explanation:

We have to take into account the expression for the position of the fringes

dsen\theta=m\lambda\\y=\frac{m\lambda D}{d}

where m is the number of the maximum, d is the separation of the slits, D is the distance to the screen.

(a) By replacing we obtain

y(m=1)=\frac{(1)(600*10^{-9}m)(1.1m)}{0.22m}=3*10^{-6}m\\\\y(m=1)=\frac{(-1)(600*10^{-9}m)(1.1m)}{0.22m}=-3*10^{-6}m

(b)  more information is required to solve this point. Please complete the information.

HOPE THIS HELPS!

4 0
3 years ago
Read 2 more answers
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