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faltersainse [42]
3 years ago
8

On a sunny day, a family decided to take sail on a nearby lake. During sailing the wind applies a force of 25N north on the sail

s of the sailboat. The
water however exerts a force of 20 N east. If the total mass of both the sailboat and the crew (family on board) is 20kg, what is the magnutude and
direction of the acceleration produced?
​
Physics
1 answer:
natka813 [3]3 years ago
4 0

Answer:

its n

Explanation:

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19. A 0.0340 kg bullet traveling at 120 m/s embeds itself in a 1.24 kg wooden block which
Blababa [14]

Answer: x ≈ 36.3 cm

Explanation:

Conservation of momentum during the collision

0.0340(120) + 1.24(0) = (0.0340 + 1.24) v

v = 3.2025 m/s

The kinetic energy of the block/bullet mass will convert to spring potential

½kx² = ½mv²

x = √(mv²/k)

x = √(1.274(3.2025²) / 99.0)

x = 0.363293... ≈ 36.3 cm

3 0
3 years ago
Ok so I am taking a test I need help please
Aloiza [94]

Answer:

i know its definetly either clockwise or counter clockwise

Explanation:

6 0
3 years ago
A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t
Yuliya22 [10]

Answer:

a) W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

b) W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

c) V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

d)  d_1 =0.183m or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

Part b

For this case first we can convert the spring constant to N/m like this:

2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}

And the work donde by the spring on this case is given by:

W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i

And if we solve for the initial velocity we got:

V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

-1/2mV^2_i = W_g + W_{spring}

And replacing we got:

-1/2mV^2_i =mg d_1 -1/2 k d^2_1

And we can put the terms like this:

\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0

If we multiply all the equation by 2 we got:

k d^2_1 -2 mg d_1 -m V^2_i =0

Now we can replace the values and we got:

200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0

200 d^2_1 -2.058 d_1 -6.3525=0

And solving the quadratic equation we got that the solution for d_1 =0.183m or 18.3 cm because the negative solution not make sense.

5 0
2 years ago
When astronauts get on a rocket and go outside Earth they float because gravity is affected by this
V125BC [204]

Answer:

Distance

Explanation:

Gravity is indirectly related to distance. This means if distance increases, gravitational pull decreases.

8 0
3 years ago
An object of mass 3 kg, moving with and sticks to an object of mass 2.00 kg with an initial velocity of 3.00 m/s. Find the final
Aleonysh [2.5K]

Answer:

1.8 m/s

Explanation:

momentum = mass × velocity

initial momentum = m1v1+m2v2

                             = 3×3 +2×0 = 9+0= 9 kg m/s

let combined velocity be V

HENCE  

final momentum = total mass × velocity

                            = (3+2) × V = 5V

According to law of conservation of momentum

final momentum = initial momentum

5V = 9

V =9/5

V = 1.8 m/s

8 0
3 years ago
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