An m = 6.11 g bullet is fired into a 365 g block that is initially at rest at the edge of a table of h = 1.08 m height as shown in the figure.
1. The bullet remains in the block, and after the impact the block lands d = 1.84 m from the bottom of the table. How much time does it take the block to reach the ground once it flies off the edge of the table?
2. What is the initial horizontal velocity of the block as it flies off the table?
3. Determine the initial speed of the bullet.

Answer:

3.75 m/s

Explanation:

Comment

Let's deal with the second question first. The average velocity is 0 because the displacement (distance between the starting point and ending point) is 0.

The first question is a little harder. You don't seem to have enough information. When that happens, you can just make things up. Now there's an interesting solution. You could do it with algebra, but it is easier to see with numbers.

Givens

Formulas

t1 = d / r1

t2 = d/r2

average rate = total distance / total time

Solution

t1 = 150 m / 5 m/s = 30 seconds

t2 = 150 m / 3 m/s = 50 seconds

Total distance = 150 m + 150 m = 300 m

Total time = 30 seconds + 50 seconds = 80 seconds

Average speed = 300 m / 80 s

Answer

Average speed = 3.75 m/s

The total distance traveled by the gardener is 8 meters. the magnitude of total displacement is approximately 7 meters.

                                                                      = 8 meters

For the magnitude of displacement, we will take the shortest path for the gardener movement.

                              for displacement-

                                       [tex]=\sqrt{( 0-4^{2}) +( 6-0^{2} )}\\=\sqrt{52} \\=7[/tex](approximately)

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The sentence that uses the subjunctive mood correctly is as follows:

Thus, the correct option is D.

Subjunctive mood may be defined as statements or actions that are in the probable situation of doubt and suspicion.

Except for the sentence fourth, the rest all indicates the factual concept with the appropriate condition.

Therefore, the correct option for this question is D.

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Answer:

C. it is important that she apologize to her sister for what she said

Explanation:

test approved

(a) The equilibrant C for force of vector A and B is 3.43 N.

(b) The equilibrant C for fx of vector A and B is 2.1 N.

(c) The equilibrant  C, for fy of vector A and B is 2.12 N.

An equilibrant force is a single force that will bring other bodies into equilibrium.

Vector A: mass = 0.2 kg, θ = 20⁰

Vector B: mass = 0.15 kg, θ = 80⁰

Fx = mg cosθ

Fy = mg sinθ

where;

Force of A due to its weight

F(A) = mg

F(A) = 0.2 x 9.8 = 1.96 N

Fx = (0.2 x 9.8) cos(20) = 1.84 N

Fy = (0.2 x 9.8) sin(20) = 0.67 N

R = √(0.67² + 1.84²)

R = 1.96 N

Force of B due to its weight

F(B) = mg

F(B) = 0.15 x 9.8

F(B) = 1.47 N

Fx = (0.15 x 9.8) cos(80) = 0.26 N

Fy = (0.15 x 9.8) sin(80) = 1.45 N

R = √(0.26² + 1.45²)

R= 1.47 N

Equilibrant force:

Force, C = 1.96 N + 1.47 N

Force, C = 3.43 N

Equilibrant FX:

Fx, C = Fx(A) + Fx(B)

Fx, C = 1.84 N + 0.26 N = 2.1 N

Equilibrant FY:

Fy, C = Fy(A) + Fy(B)

Fy, C =0.67 N + 1.45 N = 2.12 N

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A beta particle or an electron is released during beta decay. The charges can be positive or negative. Co changes to Ni, Fe to Mn, Pb to Tl, and Pu to Am.

Beta-decay is the radioactive decay that involves the release of the beta particle or the positron or the electron. The larger nucleus splits during nuclear fission and releases smaller nuclei.

The nuclear reactions are shown as:

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Answer:

The force between two objects is calculated through the equation,

                       F = Gm₁m₂/d²

where m₁ and m₂ are the masses of the objects. In this case, an unknown mass and Earth. d is the distance between them and G is the universal gravitation constant.

In the second case, if the force is to become 2.5 times the original and all the variables are constant except d then,

                      2.5F = Gm₁m₂ / (D²)

                               D = 0.623d

Subsituting the known value of d,

                               D = 0.623(6.9 x 10^8) = 4.298 x 10^8 m

Hi there!

The maximum deformation of the bumper will occur when the car is temporarily at rest after the collision. We can use the work-energy theorem to solve.

Initially, we only have kinetic energy:

[tex]KE = \frac{1}{2}mv^2[/tex]

KE = Kinetic Energy (J)
m = mass (1060 kg)
v = velocity (14.6 m/s)

Once the car is at rest and the bumper is deformed to the maximum, we only have spring-potential energy:

[tex]U_s = \frac{1}{2}kx^2[/tex]

k = Spring Constant (1.14 × 10⁷ N/m)

x = compressed distance of bumper (? m)

Since energy is conserved:

[tex]E_I = E_f\\\\KE = U_s\\\\\frac{1}{2}mv^2 = \frac{1}{2}kx^2[/tex]

We can simplify and solve for 'x'.

[tex]mv^2 = kx^2\\\\x = \sqrt{\frac{mv^2}{k}}[/tex]

Plug in the givens and solve.

[tex]x = \sqrt{\frac{(1060)(14.6^2)}{(1.14*10^7)}} = \boxed{0.0198 m}[/tex]

The work done by the force on the box 588 Joule.

It is the energy that is stored in an object due to its position relative to some zero position.

According to the question,

Given mass of the box (m)  = 6.0 kg

Given displacement of the box =  10.00 m

Hence , the height of the box  (h) = 10.00 m

Acceleration due to gravity (g)  =  9.8 m/s²

Since ,

The upward force was applied to the box, the box moves up to the height.

Therefore, the work done by the ball is the potential energy gained by the ball.

By using the formula of potential energy

P.E. = mgh

= 6 × 9.8 × 10 Joule  = 588 Joule

As we know .

Work done (W) = Potential energy (P.E.)

Work done (W) = 588 Joule

Thus,  588 Joule is the work done by the force on the box.

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The velocity of the air relative to the runner is 5 m/s.

We must recall that velocity is a vector quantity and the relative velocity must be obtained vectorially. Thus we know that;

Velocity of the runner = 4m/s. due west

Velocity of the wind =  3m/s due south

The relative velocity is;

Vr = √(4)^2 + (3)^2

Vr = 5 m/s

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(a) The time spent in the air by the ball is 1.5 seconds.

(b) The horizontal distance of the ball is 9.83 m.

(c) The impact speed of the ball is 20.37 m/s.

(2) The magnitude of the force is 253.5 N in horizontal direction.

The time of motion of the ball is calculated as follows;

h = vt + ¹/₂gt²

18 = 8sin(35)t + (0.5)(9.8)t²

18 = 4.589t + 4.9t²

4.9t² + 4.589t - 18 = 0

solve the quadratic equation using formula method,

t = 1.5 s

X = vcosθ(t)

X = 8cos(35) x 1.5

X = 9.83 m

vyf = vyi + gt

vyf = 8sin(35) + 9.8(1.5)

vyf = 19.289 m/s

vxf = vxi

vxf = 8 x cos(35)

vxf = 6.55 m/s

Resultant speed = √(19.289² + 6.55²) = 20.37 m/s

F = ma

F = 65 x 3.9

F = 253.5 N

Since the motion is horizontal, the angle of the motion is zero.

Fx = 253.5cos(0) = 253.5 N

Fy = 253.5sin(0) = 0 N

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The coefficient of friction between the block and the surface is determines as 0.1.

Apply the principle of conservation of linear momentum;

0.05(400) = v(0.05 + 25)

20 = 25.05v

v = 0.8 m/s

Apply the principle of conservation of energy

P.E  - K.E = Ux

μmgx - ¹/₂mv²  = ¹/₂kx²

μ(25.05)(9.8)(0.12) - 0.5(25.05)(0.8)² = 0.5(180)(0.12)

29.459μ - 8.016   = 10.8

29.459μ = 2.784

μ = 0.1

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Answer: the plot that you see is called a blackbody spectrum. This plot tells us the intensity or the amount of light that an object will emit at different wavelenths.

Explanation: i hope that this helps

Answer:

i don't know sorry

i know a good website that may help you

it's called https://skyandtelescope.org/observing/telescope-calculator/

Explanation:

The net force on the crate will be 99 N.Option B is correct.

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N). it is defined as the product of the coefficient of friction and normal reaction.

On resolving the given force and acceleration.Mathematically in the different components and balancing the equation gets.Components in the x-direction.

When all the forces get resolved the y-direction of forces are;

300 sin 10° +N = 445

N = 445-295.4

N = 392.9 N

The normal force is 392.9 N.

The net force on the crate is found by resolving the force in the x-direction;

F = 300 cos 10° - μN

F= 295.44-196.45

F= 98.99

F=99

Hence, the net force on the crate will be 99 N.Option B is correct.

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Answer:

suna bro i am nepali and nepali are pro

[tex]q = -21 * 10^{-6} C[/tex]

The acceleration is constant and equal to the gravitational acceleration g which is 9.8 m/s at sea level on the Earth.

As we know that charge and its sign that remains in equilibrium is under gravity must be such that it will balance the gravitational force by electric force,

mg =qE

[tex]1.45 * 10^{-3}(9.80)=q(600)\\\\q = 21 *10^{6} C\\\\[/tex]

and its sign must be negative so that it will have upward electric force

so it is

[tex]q = -21 * 10^{-6} C[/tex]

The charge of a particle of mass is [tex]-21 * 10^{-6} C[/tex]

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Answer:

i believe the answer is true

Answer:

Explanation:

The formula for Speed is distance per time.

Data

D = 40 m

T = 10 sec

S = ?

To find the speed , divide the distance by the time.

Note : the unit for Speed is meter /second

Therefore the speed of the wave, the surfer is riding on is 4 m/s.

Gravity and normal forces act on the toy car.

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A 3rd harmonic of a tube open at both ends will have displacement antinodes at both ends.

In a tube of length L with two open ends, the longest standing wave has displacement antinodes (pressure nodes) at both ends. The fundamental or first harmonic is what it is known as. The second harmonic is the longest standing wave in a tube of length L with two open ends.

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Answer:

di

Explanation:

The prefix is di. CO2 is carbon dioxide. di means 2.

Answer:

[tex]\textsf {v = 20 m/s}[/tex]

[tex]\mathsf {a = 16 m/s^2}[/tex]

Explanation:

[tex]\textsf {To find velocity, let's differentatiate x :}[/tex] [tex]\implies \mathsf {v = \frac{dx}{dt} }[/tex]

⇒ [tex]\mathsf {v = \frac{d}{dx}(t^{3}+2t^{2}+4) }[/tex]

⇒ [tex]\mathsf {v = 3t^{2} +4t +0 }[/tex] [tex]\textsf {(By applying law of differentiation)}[/tex]

[tex]\textsf {Substitute t = 2 :}[/tex]

⇒ [tex]\mathsf {v = 3(2)^{2} + 4(2) }[/tex]

⇒ [tex]\textsf {v = 12 + 8}[/tex]

⇒ [tex]\textsf {v = 20 m/s}[/tex]

[tex]\textsf {To find acceleration, let's differentatiate v :}[/tex] [tex]\implies \mathsf {a = \frac{dv}{dt} }[/tex]

⇒ [tex]\mathsf {a = \frac{d}{dx}(3t^{2}+4t) }[/tex]

⇒ [tex]\textsf {a = 6t + 4}[/tex]

[tex]\textsf {Substitute t = 2 :}[/tex]

⇒ [tex]\textsf {a = 6(2) + 4}[/tex]

⇒ [tex]\textsf {a = 12 + 4}[/tex]

⇒ [tex]\mathsf {a = 16 m/s^2}[/tex]

The distance and speed are as follows:

Speed of a body is the ratio of the distance covered by a body and the time it takes to cover that distance,

The number of seconds in a year = 365.25 * 24 * 3600 = 31557600 second

a) Distance covered in 51 seconds = 5.5/31557600 * 51

Distance = 8.89 * 10⁻⁶ cm

b) 5.5 cm = 3.4175 * 10⁻⁵ miles

Sped in miles per million years = 3.42 * 10⁻⁵ miles/1 * 10⁻⁶ million years

Speed = 34.2 miles per million years.

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The distance covered by an accelerating object starting from rest is

D = (1/2 of acceleration) x (time²).

In this question,

D = 100 m

time = 3.8 s

100 = (1/2 acceleration) x (3.8s²)

100 = (1/2 a) x (14.44)

100/14.44  =  1/2 a

200/14.44  =  a

a = 13.85 m/s²

The work function in eV for the given cutoff frequency is  5.05 eV.

The work function is related to the frequency as

Wo = h x fo

where, fo = cutoff frequency and h is the Planck's constant

Given is the cutoff frequency for a certain element is 1.22 x 10¹⁵ Hz

Wo = 6.626 x 10⁻³⁴ x  1.22 x 10¹⁵ Hz / 1.6 x 10⁻¹⁹

Wo = 5.05 eV

Thus, the work function is  5.05 eV

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The initial speed of the object with mass "m" is 4.79 m/s and the initial speed of the object with mass "2m" is 1.81 m/s.

The initial speed of the two masses is calculated by applying the principle of conservation of linear momentum as follows;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

[tex]m_1u_1sin(\theta_1) + m_2u_2sin(\theta_2) = (m_1 + m_2)vsin(\theta_3)\\\\mu_1sin(0) + 2mu_2sin(90) = (m + 2m)(2)sin(37)\\\\2mu_2 = 3.61m\\\\u_2 = \frac{3.61 m}{2m} \\\\u_2 = 1.81 \ m/s[/tex]

[tex]m_1u_1sin(\theta_1) + m_2u_2sin(\theta_2) = (m_1 + m_2)vsin(\theta_3)\\\\mu_1cos(0) + 2mu_2cos(90) = (m + 2m)(2)cos(37)\\\\mu_1 = 4.79 m\\\\u_1 = 4.79 \ m/s[/tex]

Thus, the initial speed of the object with mass "m" is 4.79 m/s and the initial speed of the object with mass "2m" is 1.81 m/s.

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The acceleration of Karla's I IPhone being thrown from Mr. Higley's classroom at 0 m/s will be 29.16 m/s²

The rate of velocity change concerning time is known as acceleration. According to Newton's second law, the eventual effect of all forces applied to a body is its acceleration.

The pace at which a body's velocity varies is represented by acceleration, which is a vector quantity.

The given data in the problem is given by ;

u is the initial speed =  0 m/sec

v is the final speed= 35  m/sec

t is the time interval= 1.2 second

a is the acceleration=? m/sec²

The formula for acceleration is;

[tex]\rm a=\frac{v-u}{t} \\\\ a= \frac{35-0}{1.2} \\\\ a= 29.16 \ m./s^2[/tex]

Hence, the acceleration of Karla's iPhone being thrown from Mr. Higley's classroom at 0 m/s will be 29.16 m/s²

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The friction force is 1.5N

Now, using the third equation of motion-

[tex]v^{2}= u^{2} +2as[/tex]

v=final velocity

u=initial velocity

a=acceleration

s=distance

substitute the values in the above equation

1.2=0+ 2a×20

a=1.2/40

a=0.03

friction force=mass ×acceleration

                    =50×0.03

                    =1.5N

hence, The friction force is 1.5N

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