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DENIUS [597]
3 years ago
8

If the acceleration of a motorboat is 4.0 m/s2, and the motorboat starts from rest, what is its velocity after 6.0 s?

Physics
1 answer:
umka2103 [35]3 years ago
6 0

Answer:

The velocity of the motorboat after 6s is 24 m/s.

Explanation:

Given;

acceleration of the motorboat, a = 4.0 m/s²

initial velocity of the motorboat, u = 0

time of motion of the motorboat = 6s

Apply the following kinematic equation to determine the velocity of the motorboat after 6 ;

v = u + at

v = 0 + (4 x 6)

v = 24 m/s

Therefore, the velocity of the motorboat after 6s is 24 m/s.

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a concrete slab 20 m long and weighing 400,000 N is supported by one pillar. A 19,600 N car is parked 8 meters from one end, whe
Elden [556K]

let the distance of pillar is "r" from one end of the slab

So here net torque must be balance with respect to pillar to be in balanced state

So here we will have

Mg(r - L/2) = mg(L/2 - 8)

here we know that

mg = 19600 N

Mg = 400,000 N

L = 20 m

from above equation we have

400,000(r - 10) = 19,600 (10 - 8)

r - 10 = 0.098

r = 10.098 m

so pillar is at distance 10.098 m from one end of the slab

3 0
3 years ago
A bullet with an initial kinetic energy of 400 J strikes a wooden block where a 8000 N resistive force stops the bullet. What is
Sindrei [870]

Answer:

d = 0.05 [m] = 50 [mm]

Explanation:

We must remember the principle of conservation of energy which tells us that energy is transformed from one way to another. For this case, the initial kinetic energy is transformed into useful work that is equal to the product of force by distance.

E_{k}=F*d\\400 = 8000*d\\d = 0.05 [m] = 50 [mm]

5 0
2 years ago
What is the total force acting on the ball in the diagram? How would you describe its velocity?
KonstantinChe [14]

Answer:

Velocity formulae should be used in order to know the answer

8 0
3 years ago
Read 2 more answers
A yo‑yo with a mass of 0.0800 kg and a rolling radius of =2.70 cm rolls down a string with a linear acceleration of 5.70 m/s2.
N76 [4]

Explanation:

Given that,

Mass, m = 0.08 kg

Radius of the path, r = 2.7 cm = 0.027 m

The linear acceleration of a yo-yo, a = 5.7 m/s²

We need to find the tension magnitude in the string and the angular acceleration magnitude of the yo‑yo.

(a) Tension :

The net force acting on the string is :

ma=mg-T

T=m(g-a)

Putting all the values,

T = 0.08(9.8-5.7)

= 0.328 N

(b) Angular acceleration,

The relation between the angular and linear acceleration is given by :

\alpha =\dfrac{a}{r}\\\\\alpha =\dfrac{5.7}{0.027}\\\\=211.12\ m/s^2

(c) Moment of inertia :

The net torque acting on it is, \tau=I\alpha, I is the moment of inertia

Also, \tau=Fr

So,

I\alpha =Fr\\\\I=\dfrac{Fr}{\alpha }\\\\I=\dfrac{0.328\times 0.027}{211.12}\\\\=4.19\times 10^{-5}\ kg-m^2

Hence, this is the required solution.

3 0
3 years ago
You want to photograph a circular diffraction pattern whose central maximum has a diameter of 1.0 cm. You have a helium-neon las
topjm [15]

Answer:

0.776 m far Pinhole should be placed before the viewing screen

Explanation:

For circular aperture of diameter D will have a bright central maximum of diameter, width is given by

w=\frac{2.44 \lambda L}{D}

where \lambda is wavelength of helium neon laser = 633 nm, D=10.cm, w=0.12 mm

Pinhole should be placed before the viewing screen is

L=\frac{wD}{2.44\lambda}\\L=\frac{0.12\times 10^{-3}\times 0 .01}{2.44\times 633 \times 10^{-9}}\\L=0.776 m

4 0
3 years ago
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