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klio [65]
3 years ago
8

The diagram shows a ballistic pendulum. A 200 g bullet is fired into the suspended 4 kg block of wood and remains embedded insid

e it (a perfectly inelastic collision). After the impact of the bullet, the block swings up to a maximum height h. The initial speed of the bullet was 50 m/s.
Physics
1 answer:
adoni [48]3 years ago
5 0

Question

What was the initial momentum of the bullet before collision?

Answer:

10 Kg.m/s

Explanation:

Momentum is a product of velocity of an object in m/s and its mass in kgs hence numerically expressed as p=mv where p is momentum, v is velocity and m is mass. Substituting m for 0.2 kg and v for 50 m/s then p=0.2*50=10 kg.m/s

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Gravity is dependent on which of the two factors?
k0ka [10]
Gravity is dependent on Mass & Distance

7 0
3 years ago
A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's in
german

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

\Sigma F = -f = m\cdot a (Eq. 1)

Where:

f - Kinetic friction force, measured in newtons.

m - Mass of the box, measured in kilograms.

a - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight (W = m\cdot g) and uniform accelerated motion (v = v_{o}+a\cdot t), we expand the previous expression:

-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)

And the magnitude of the friction force exerted on the box is calculated by this formula:

f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right) (Eq. 1b)

Where:

W - Weight, measured in newtons.

g - Gravitational acceleration, measured in meters per square second.

v_{o} - Initial speed, measured in meters per second.

v - Final speed, measured in meters per second.

t - Time, measured in seconds.

If we know that W = 52.4\,N, g = 9.807\,\frac{m}{s^{2}}, v_{o} = 1.37\,\frac{m}{s}, v = 0\,\frac{m}{s} and t = 2.8\,s, the magnitud of the kinetic friction force exerted on the box is:

f = -\left(\frac{52.4\,N}{9.807\,\frac{m}{s^{2}} } \right)\cdot \left(\frac{0\,\frac{m}{s}-1.37\,\frac{m}{s}  }{2.8\,s} \right)

f = 2.614\,N

The magnitude of the friction force exerted on the box is 2.614 newtons.

5 0
3 years ago
Projectile's horizontal range on level ground is R=v20sin2θ/g. At what launch angle or angles will the projectile land at half o
seraphim [82]

Answer:

\theta = 15^o \: or\: 75^o

Explanation:

As we know that the formula of range is given as

R = \frac{v^2sin2\theta}{g}

now we know that

maximum value of the range of the projectile is given as

R_{max} = \frac{v^2}{g}

now we need to find such angles for which the range is half the maximum value

so we will have

\frac{R}{2} = \frac{v^2}{2g} = \frac{v^2sin(2\theta)}{g}

sin(2\theta) = \frac{1}{2}

2\theta = 30 or 150

\theta = 15^o \: or\: 75^o

7 0
3 years ago
An accepted value for the acceleration due to gravity is 9.801 m/s2. In an experiment with pendulums, you calculate that the val
Fed [463]

g Generally the accepted value of acceleration due to gravity is 9.801 m/s^2

as per the question the acceleration due to gravity is found to be 9.42m/s^2 in an experiment performed.

the difference between the ideal and observed value is 0.381.

hence the error is -\frac{0.381}{9.801} *100

                                                            =3.88735 percent

the error is not so high,so it can be  accepted.

now we have to know why this occurs-the equation of time period of the simple pendulum is give as-T=2\pi\sqrt[2]{l/g}

                                                      g=4\pi^2\frac{l}{T^2}

As the experiment is done under air resistance,so it will affect to the time period.hence the time period will be more which in turn decreases the value of g.

if this experiment is done in a environment of zero air resistance,we will get the value of g which must be approximately equal to 9.801  m/s^2

5 0
3 years ago
If Fg=mg solve for g
Bas_tet [7]

Answer:Fg = mg however newtons second law states that the net force acting on an object is equal to it's mass times it's acceleration so what allows us to say that Fg = mg because certainly not for every single situation the net force is going to equal to the force of gravity please explain... what allows us to say Fg = mg

Source https://www.physicsforums.com/threads/fg-mg-questioned.336776/

Explanation:

6 0
3 years ago
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