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skad [1K]
2 years ago
8

A car with a velocity of 15 m/s is accelerated uniformly at the rate of 1.9 m/s2 for 6.8 s. What is its final velocity?

Physics
1 answer:
creativ13 [48]2 years ago
3 0

Answer:

33 m/s

Explanation:

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radiation

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2 years ago
The law of reflection states that if the angle of incidence is 39 degrees, the angle of reflection is ___ degrees.
marishachu [46]
It's 39 degrees, too


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8 0
3 years ago
A 3.9 g dart is fired into a block of wood with a mass of 24.6 g. The wood block is initially at rest on a 1.5 m tall post. Afte
Galina-37 [17]

Answer:

46.48m/s

Explanation:

The problem is a combination of the principle of conservation of linear momentum and projectile motion.

The principle of conservation of linear momentum states that in a closed system, the total momentum of colliding bodies before impact is equal to the total momentum after impact. The masses stated in the problem experienced an inelastic collision. In an inelastic collision, the bodies involved stick together after the collision and move with a common velocity.

For two bodies of masses m_1 and m_2 moving with velocities u_1 and u_2 before impact, if they experience inelastic collision, the conservation of their momenta is as stated in equation (1);

m_1u_1+m_2u_2=(m_1+m_2)v..................(1)

were v is their common velocity after impact. If the second mass m_2 was at rest before the impact, then its initial velocity u_2=0m/s. therefore m_2u_2=0. Equation (1) then becomes;

m_1u_1=(m_1+m_2)v..............(2)

In the problem stated, the second mass taken as the mass of the wooden block was at rest before the impact and the collision was inelastic since both the wood and the dart stuck together and moved with a common velocity after the impact. Therefore we can use equation (2) for the problem.

Given;

m_1=3.9g=0.0039kg\\u_1=?\\m_2=24.6g=0.0246kg\\v=?

Substituting these values into (2), we get the following;

0.0039*u_1=(0.0039+0.024)v\\0.0039u_1=0.0285v.........(3)

Their common v velocity after impact now makes both the wooden block and the dart (as a single body) to fall vertically through a height h of 1.5m over a range R of 3.5m as stated by the problem; hence by the principle of projectile motion for a body projected horizontally, the following relationship holds;

R= vt............(4)

were t is the time taken to fall through the height h. To obtain t we use the second equation of free fall under gravity;

h=\frac{1}{2}gt^2...........(5)

were g is acceleration due to gravity taken as 9.8m/s^2. Therefore;

1.5=\frac{1}{2}*9.8*t^2\\1.5=4.9t^2\\t^2=\frac{1.5}{4.9}=0.306\\t=\sqrt{0.306} =0.55s

We then substitute R and t into equation (4) to obtain v.

3.5=v*0.55\\v=\frac{3.5}{0.55}\\v=6.36m/s

We now further substitute this value of v into (3) to obtain u_1;

u_1=\frac{0.0285v}{0.0039}\\\\u_1=\frac{0.0285*6.36}{0.0039}\\\\u_1=\frac{0.18126}{0.0039}\\\\u_1=46.48m/s

4 0
3 years ago
The moment of inertia of a uniform-density disk rotating about an axle through its center can be shown to be . This result is ob
Naddik [55]

(a) 0.2888 kg m^2

The moment of inertia of a uniform-density disk is given by

I=\frac{1}{2}MR^2

where

M is the mass of the disk

R is its radius

In this problem,

M = 16 kg is the mass of the disk

R = 0.19 m is the radius

Substituting into the equation, we find

I=\frac{1}{2}(16 kg)(0.19 m)^2=0.2888 kg m^2

(b) 142.5 J

The rotational kinetic energy of the disk is given by

K=\frac{1}{2}I\omega^2

where

I is the moment of inertia

\omega is the angular velocity

We know that the disk makes one complete rotation in T=0.2 s (so, this is the period). Therefore, its angular velocity is

\omega=\frac{2\pi}{T}=\frac{2\pi}{0.2 s}=31.4 rad/s

And so, the rotational kinetic energy is

K=\frac{1}{2}(0.2888 kg m^2)(31.4 rad/s)^2=142.5 J

(c) 9.07 kg m^2 /s

The rotational angular momentum of the disk is given by

L=I\omega

where

I is the moment of inertia

\omega is the angular velocity

Substituting the values found in the previous parts of the problem, we find

L=(0.2888 kg m^2)(31.4 rad/s)=9.07 kg m^2 /s

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3 years ago
A tsunami covers a distance of 500km in 83.3 minutes. If the period of vibration of the molecules is 35.0 minutes, what is the w
marin [14]

Answer:

210079.798 m

Explanation:

pls give brainliest :)

8 0
2 years ago
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