Ask question

Ask question

All categories

3 years ago

13

A dart gun suspended by strings hangs in equilibrium. The mass of the gun is 355 grams, not including a dart. The gun fires a 57

.0 gram dart, causing it to swing backwards. The gun swings up to a height of 18.3 centimeters. What was the dart's speed in meters per second just after firing?

1 answer:

gogolik [260]3 years ago

5 0

Answer:

dart's speed is 11.77 m/s

Explanation:

given data

mass of the gun m = 355 gram = 0.355 kg

height = 18.3 centimeters = 0.183 m

dart = 57.0 gram = 0.057 kg

to find out

dart's speed

solution

we apply here law of conservation of energy that is express as

mgh = 0.5 × m × v² ...........1

so speed of gun will be here as

V = $\sqrt{2gh}$ ..................2

V = $\sqrt{2*9.8*0.183}$

V = 1.89 m/s

and

now we find speed of dart by use law of conservation of momentum that is

M×V = m×v ...............3

so speed of the dart is

v = $\frac{M*V}{m}$

v = $\frac{0.355*1.89}{0.057}$

v = 11.77 m/s

so dart's speed is 11.77 m/s

You might be interested in

Three equal charges of magnitude 'e' are located at the vertices of an equilateral triangle of side 1m. Where should you place a

Naily [24]

Answer:

At centroid

Explanation:

In the given equilateral triangle ABC with side of 1 m. The three equal charges e,e,e are placed at the A,B and C.

And the fourth charge 2e is put at point O which is called centroid.

Now we can calculate the distance AD by applying pythagorean theorem as,

$AD^{2}=AB^{2}+BD^{2}$

Put the values and get.

$AD^{2}=1^{2}+(\frac{1}{2} )^{2}\\AD=\sqrt{\frac{3}{4} } \\AD=\frac{\sqrt{3}}{2}$

Now calculate AO as,

$AO=\frac{2}{3}\times \frac{\sqrt{3} }{2}\\AO=\frac{1}{\sqrt{3} }$

And the sides BO=CO=AO.

Now Force can be calculated as

$F_{1}=\frac{2ke^{2} }{\frac{1}{\sqrt{3} } ^{2} }\\F_{1}=6ke^{2}$

And similarly,

$F_{2}=F_{3}=6ke^{2}$

Now we can calculate resultant of $F_{2}andF_{3}$ in upward direction. as,

$F_{net}=\sqrt{F_{2}^{2}+F_{3}^{2}+2F_{2}F_{3}cos120 } \\F_{net}=\sqrt{F_{2}^{2}+F_{2}^{2}+2F_{2}F_{2}(-\frac{1}{2})}\\F_{net}=6ke^{2}$

Therefore the resultant force on centroid O.

$F=F_{1}-F_{net}\\F=6ke^{2}-6ke^{2}\\F=0$

Therefore the fourth charge 2e should be placed on centroid so that it experience zero force.

8 0

2 years ago

What is the kinetic energy of a 14kg object traveling at 10m/s

Sauron [17]

Answer:

Explanation:

kinetic energy=1/2*mass*velocity^2

=1/2*14kg*10^2

=7*100

=700 joule

5 0

2 years ago

A 2.964 kilogram truck strikes a fence at 7.00 meters/second and comes to

Makovka662 [10]

10.92N

Explanation:

Given parameters:

Mass of truck = 2.964kg

Velocity of truck = 7m/s

Time taken = 1.9s

Unknown:

Average force on the car = ?

Solution:

According to newton's third law of motion "action and reaction are equal and opposite".

The force with which the truck struck the fence is the same as the force the fence acted on the truck with but in another direction.

From newton's second law:

Force = mass x acceleration

We know that acceleration is the change in velocity with time;

acceleration = $\frac{change in velocity }{time }$

Force = mass x $\frac{change in velocity }{time }$

Force = $\frac{2.964 x 7}{1.9}$ = 10.92N

learn more:

Newton's laws brainly.com/question/11411375

#learnwithBrainly

3 0

2 years ago

A 1 kg mass is attached to a spring with spring constant 7 Nt/m. What is the frequency of the simple harmonic motion? What is th

Scorpion4ik [409]

1. 0.42 Hz

The frequency of a simple harmonic motion for a spring is given by:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

where

k = 7 N/m is the spring constant

m = 1 kg is the mass attached to the spring

Substituting these numbers into the formula, we find

$f=\frac{1}{2\pi}\sqrt{\frac{7 N/m}{1 kg}}=0.42 Hz$

2. 2.38 s

The period of the harmonic motion is equal to the reciprocal of the frequency:

$T=\frac{1}{f}$

where f = 0.42 Hz is the frequency. Substituting into the formula, we find

$T=\frac{1}{0.42 Hz}=2.38 s$

3. 0.4 m

The amplitude in a simple harmonic motion corresponds to the maximum displacement of the mass-spring system. In this case, the mass is initially displaced by 0.4 m: this means that during its oscillation later, the displacement cannot be larger than this value (otherwise energy conservation would be violated). Therefore, this represents the maximum displacement of the mass-spring system, so it corresponds to the amplitude.

4. 0.19 m

We can solve this part of the problem by using the law of conservation of energy. In fact:

- When the mass is released from equilibrium position, the compression/stretching of the spring is zero: $x=0$ , so the elastic potential energy is zero, and all the mechanical energy of the system is just equal to the kinetic energy of the mass:

$E=K=\frac{1}{2}mv^2$

where m = 1 kg and v = 0.5 m/s is the initial velocity of the mass

- When the spring reaches the maximum compression/stretching (x=A=amplitude), the velocity of the system is zero, so the kinetic energy is zero, and all the mechanical energy is just elastic potential energy:

$E=U=\frac{1}{2}kA^2$

Since the total energy must be conserved, we have:

$\frac{1}{2}mv^2 = \frac{1}{2}kA^2\\A=\sqrt{\frac{m}{k}}v=\sqrt{\frac{1 kg}{7 N/m}}(0.5 m/s)=0.19 m$

5. Amplitude of the motion: 0.44 m

We can use again the law of conservation of energy.

- $E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$ is the initial mechanical energy of the system, with $x_0=0.4 m$ being the initial displacement of the mass and $v_0=0.5 m/s$ being the initial velocity

$- E_f = \frac{1}{2}kA^2$ is the mechanical energy of the system when x=A (maximum displacement)

Equalizing the two expressions, we can solve to find A, the amplitude:

$\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}kA^2\\A=\sqrt{x_0^2+\frac{m}{k}v_0^2}=\sqrt{(0.4 m)^2+\frac{1 kg}{7 N/m}(0.5 m/s)^2}=0.44 m$

6. Maximum velocity: 1.17 m/s

We can use again the law of conservation of energy.

- $E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$ is the initial mechanical energy of the system, with $x_0=0.4 m$ being the initial displacement of the mass and $v_0=0.5 m/s$ being the initial velocity

$- E_f = \frac{1}{2}mv_{max}^2$ is the mechanical energy of the system when x=0, which is when the system has maximum velocity, $v_{max}$

Equalizing the two expressions, we can solve to find $v_{max}$ , the maximum velocity:

$\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}mv_{max}^2\\v_{max}=\sqrt{\frac{k}{m}x_0^2+v_0^2}=\sqrt{\frac{7 N/m}{1 kg}(0.4 m)^2+(0.5 m/s)^2}=1.17 m/s m$

4 0

2 years ago

Read 2 more answers

A ball is falling after rolling off a tall roof. The ball has

Rom4ik [11]

C.
Because it’s falling it has acceleration in the y direction. If you have acceleration, you usually also have velocity, and since kinetic energy is KE= Mv^2 you know you have it. It also has potential energy because it has some height to it, and PE= Mgh.

8 0

3 years ago

Read 2 more answers