A cart is pulled by a force of 250 N at an angle of 35° above the horizontal. The cart accelerates at 1.4 m/s2. The free-body diagram shows the forces acting on the cart.

A free body diagram with 3 force vectors. The first vector is pointing downward, labeled F Subscript g Baseline. The second vector is pointing up, labeled F Subscript N Baseline. The third vector is pointing up to the right at an angle of 35 degrees, labeled F Subscript p Baseline = 250 N. The up and down vectors are the same length.

The mass of the cart, to the nearest whole number, is

Answer:

Violet light has a wavelength of ~400 nm, and a frequency of ~7.5*1014 Hz.

Explanation:

Answer:

F = 35651 [N]

Explanation:

To solve this problem we must use the following equation of kinematics

[tex]v_{f} ^{2} =v_{o} ^{2} +2*a*x[/tex]

where:

Vf = final velocity = 185 [m/s]

Vo = initial velocity = 0 (starts from the rest)

a = acceleration [m/s²]

x = distance = 3.6 [m]

Now we can find the acceleration.

185² = 0 + 2*a*3.6

34225 = 7.2*a

a = 4753.47 [m/s²]

Now we must use Newton's second law which tells us that the total force on a body is equal to the product of mass by acceleration.

F = m*a

where:

m = mass = 7.5 [kg]

F = force [N] (units of Newtons)

F = 7.5*4753.47

F = 35651 [N]

Answer:

1.358 10e8 have good day please mark brainliest

Answer:

40N

160N

360N

Explanation:

The drag force can be calculated using the formula below

FD = 0.5CPAV²

The drag coefficient = c = 0.8

The frontal density = p = 1.0kg

The frontal surface area = A = 1.0

The velocity v = 10m/s, 20m/s, 30m/s

When V = 10m/s

Fd = 0.5(0.8)(1)(1)(10)²

Fd = 40N

When V = 20m/s

FD = 0.5(0.8)(1)(1)(20)²

FD = 160N

When V = 30m/s

FD = 0.5(0.8)(1)(1)(30)²

FD = 360N

Thank you

Answer:

Explanation:

Step one:

Given

mass m= 0.32kg

intital velocity, u= 14m/s

final velocity v= 22m/s

time= 0.06s

Step two:

Required

Force F

the expression for the force is

F=mΔv/t

F=0.32*(22-14)/0.06

F=(0.32*8)/0.06

F=2.56/0.06

F=42.67N

The average force exerted on the bat 42.67N

Answer:

a rubber band

Explanation:

hope this helps!

Answer:

a rubber band

Explanation:

thank you

Answer:

7.5 m

Explanation:

k = Spring constant = 180 N/m

x = Displacement of spring = 14 cm

m = Mass of projectile = 0.024 kg

a = g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

s = Displacement of projectile

v = Final velocity = 0

u = Initial velocity

The potential energy of the spring will be equal to the kinetic energy of the object

[tex]\dfrac{1}{2}kx^2=\dfrac{1}{2}mu^2\\\Rightarrow u=\sqrt{\dfrac{kx^2}{m}}\\\Rightarrow u=\sqrt{\dfrac{180\times 0.14^2}{0.024}}\\\Rightarrow u=12.12\ \text{m/s}[/tex]

[tex]v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0-12.12^2}{2\times -9.81}\\\Rightarrow s=7.5\ \text{m}[/tex]

The maximum height reached above the initial position is 7.5 m

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value is  [tex]F_{centripetal} = 15 \ N[/tex]

Explanation:

From the diagram we see that

 The tension force on the ball is  [tex]F_{Tension} = 25 \ N[/tex]

  The gravitational force on the ball is  [tex]F_{Gravity } = 10 \ N[/tex]

Generally the net centripetal force exerted on the ball is mathematically represented as

        [tex]F_{centripetal} = F_{Tension} - F_{Gravity}[/tex]

=>     [tex]F_{centripetal} = 25 - 10[/tex]

=>     [tex]F_{centripetal} = 15 \ N[/tex]

The net centripetal force exerted on the ball is 15 Newton.

Hence, Option B) 15N is the correct answer.

Given the data in the question;

Net centripetal force exerted on the ball; [tex]F_{centripetal} = \ ?[/tex]

From the diagram below, net force acting on the ball gives rise to centripetal force. Hence

[tex]F{net} = F_1 + f_2 = F_{centripetal}[/tex]

Now, from the diagram, force acting towards the center of the circular track is is positive while force directly pointing away from center is negative.

Hence

[tex]F{net} = F_1 + f_2 = F_{centripetal} = F_{Tension} - F_{Gravity}[/tex]

We substitute in our given values

[tex]F_{centripetal} = 25N - 10N\\\\ F_{centripetal} = 15N[/tex]

The net centripetal force exerted on the ball is 15 Newton.

Hence, Option B) 15N is the correct answer.

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Answer:

23 m/s downward

__________________________________________________________

Taking the downward direction as positive

We are given:

Initial velocity of the marble (u) = 0 m/s

Time interval (t) = 2.3 seconds

Final velocity (v) = x m/s

Solving for the Final velocity:

Acceleration of the Marble:

We know that gravity will make the marble accelerate at a constant acceleration of 10 m/s

Final velocity:

v = u + at                                              [First equation of motion]

x = 0 + (10)(2.3)                                    [replacing the given values]

x = 23 m/s

Hence, after 2.3 seconds, the marble will move at a velocity of 23 m/s in the downward direction

Answer:

I = 0.25 [amp]

Explanation:

To solve this problem we must use ohm's law which tells us that the voltage is equal to the product of the current by the resistance.

V = I*R

where:

V = voltage [Volt]

I = amperage or current [amp]

R = resistance [ohm]

Since all resistors are connected in series, the total resistance will be equal to the arithmetic sum of all resistors.

Rt = 2 + 8 + 14

Rt = 24 [ohm]

Now clearing I for amperage

I = V/Rt

I = 6/24

I = 0.25 [amp]

Answer: 0.25

Explanation:

Answer:

answer is B (CROWN ME)

Explanation:

One should revise his/ her research questions very well in the week before the exam. They should contribute at least 15-20 hours for going through these questions in that particular week. They must put some extra efforts on weekends.

A freely falling object near the Earth's surface has a constant: B. acceleration of 9.8 [tex]m/s^2[/tex].

Gravity can be defined as a force that controls the movement of the planets such as Earth around the Sun, hold stars grouped in galaxies together, and galaxies grouped in clusters.  This ultimately implies that, gravity is a universal force of attraction that acts between all objects that are having both mass and energy, and can occupy space.

Furthermore, the gravity near the Earth's surface makes it possible for all physical objects to possess weight and experience a free fall.

Generally, the acceleration due to gravity for an object experiencing a free fall near the Earth's surface is 9.8 [tex]m/s^2[/tex].

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Answer:

7

Explanation

There are currently seven complete periods in the periodic table, comprising the 118 known elements. Any new elements will be placed into an eighth period; see extended periodic table.

Answer:

1. At constant speed, change in the kinetic energy is zero but climbing a    mountain produces change in the potential energy

Explanation:

According to work-energy theorem, work done in moving an object from one point to another is equal to change in mechanical energy ( kinetic energy or potential energy ) of the object.

Kinetic energy is given by;

K.E = ¹/₂m(Δv)²

where;

Δv is change in speed

at constant speed, Δv = 0

Potential energy is given by;

P.E = mgΔh

where;

Δh is change in height,

there is change in height in climbing a mountain

Therefore, the best explanation in the given options is "1".

"At constant speed, change in the kinetic energy is zero but climbing a mountain produces change in the potential energy"

Answer:

The pressure at the top of the step is 129.303 kilopascals.

Explanation:

From Hydrostatics we find that the pressure difference between extremes of the water column is defined by the following formula, which is a particular case of the Bernoulli's Principle ([tex]v_{bottom}\approx v_{top}[/tex]):

[tex]p_{bottom}-p_{top} = \rho\cdot g\cdot \Delta h[/tex] (1)

[tex]p_{bottom}[/tex], [tex]p_{top}[/tex] - Total pressures at the bottom and at the top, measured in pascals.

[tex]\rho[/tex] - Density of the water, measured in kilograms per cubic meter.

[tex]\Delta h[/tex] - Height difference of the step, measured in meters.

If we know that [tex]p_{bottom} = 132000\,Pa[/tex], [tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\Delta h = 0.275\,m[/tex], then the pressure at the top of the step is:

[tex]p_{top} = p_{bottom}-\rho\cdot g\cdot \Delta h[/tex]

[tex]p_{top} = 132000\,Pa-\left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.275\,m)[/tex]

[tex]p_{top} = 129303.075\,Pa[/tex]

[tex]p_{top} = 129.303\,kPa[/tex]

The pressure at the top of the step is 129.303 kilopascals.

The pressure of the water at the top of the step is 1.293 x 10⁵ Pa.

The given parameters:

The pressure of the water at the top of the step is calculated as follows;

[tex]P_b - P_t = \rho gh\\\\P_t = P_b - \rho gh[/tex]

where;

[tex]\rho[/tex] is the density of the water

[tex]P_t = (132,000 \ Pa) \ - \ (1000 \times 9.8 \times 0.275)\\\\P_t = 1.293 \times 10^5 \ Pa[/tex]

Thus, the pressure of the water at the top of the step is 1.293 x 10⁵ Pa.

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Answer:

the coefficient of Kinetic friction between the tires and road is 0.38

Option A) .38 is the correct answer

Explanation:

Given that;

final velocity v = 0

initial velocity u = 15m/s

time taken t = 4 s

acceleration  a = ?

from the equation of motion        

v   =   u   +   at

we substitute

0 = 15 + a × 4

acceleration a = -15/4 =  - 3.75 m/s²    

the negative sign tells us that its a  deacceleration so the sign can be ignored.

Deacceleration due to friction a = μ × g

we substitute

3.75 = μ × 9.8    

μ = 3.75 / 9.8 = 0.3826 ≈ 0.38

Therefore the coefficient of Kinetic friction between the tires and road is 0.38

Option A) .38 is the correct answer

Answer:

(a) t = 5.66 s

(b) t = 8 s

Explanation:

(a)

Here we will use 2nd equation of motion for angular motion:

θ = ωi t + (1/2)∝t²

where,

θ = Angular Displacement = (3.7 rev)(2π rad/1 rev) = 23.25 rad

ωi = initial angular speed = 0 rad/s

t = time = ?

∝ = angular acceleration = 1.45 rad/s²

Therefore,

23.25 rad = (0 rad/s)(t) + (1/2)(1.45 rad/s²)t²

t² = (23.25 rad)(2)/(1.45 rad/s²)

t = √(32.06 s²)

t = 5.66 s

(b)For next 3.7 rev

θ = ωi t + (1/2)∝t²

where,

θ = Angular Displacement = (3.7 rev + 3.7 rev)(2π rad/1 rev) = 46.5 rad

ωi = initial angular speed = 0 rad/s

t = time = ?

∝ = angular acceleration = 1.45 rad/s²

Therefore,

46.5 rad = (0 rad/s)(t) + (1/2)(1.45 rad/s²)t²

t² = (46.5 rad)(2)/(1.45 rad/s²)

t = √(64.13 s²)

t = 8 s

The time taken in each case is  2.55 s

Let us recall that the equations of circular motion are almost like those of linear motion;

α = ω2 - ω1/t

α = angular acceleration

ω2 = final angular velocity

ω1 = initial angular velocity

t = time taken

a)  1.45 rad/s2 =3.70 rev/s - 0 rev/s/t

t = 3.70 rev/s/1.45 rad/s2

t = 2.55 s

b) For, ) the next 3.70 rev:

1.45 rad/s2 =3.70 rev/s - 0 rev/s/t

t = 3.70 rev/s/1.45 rad/s2

t = 2.55 s

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Explanation:

That must be the right ans.

Answer:

f = 5.65 Hz

Explanation:

The fundamental frequency of a string is given by following formula:

f = v/2L

where,

f = fundamental frequency

v = speed of wave = √(TL/m)

L = Length of String

m = Mass of String

T = Tension in String

Therefore,

f = √(TL/m)/2L

2f = √(T/Lm)

For initial condition:

T₁ = 600 N

f₁ = 110 Hz

2(110 Hz) = √(600 N/Lm)

√(600 N)/220 Hz =  √Lm

Lm = 0.01239 N/s²

Now, for changed tension:

2f₂ = √(540 N/0.01239 N/s²)

f₂ = 208.7 Hz/2

f₂ = 104.35 Hz

So, the beat frequency will be:

f = f₁ - f₂

f = 110 Hz - 104.35 Hz

f = 5.65 Hz

Answer:

I think it's c .... fulcrum

Answer:

Approximately [tex]4.39 \; \rm m \cdot s^{-1}[/tex] (assuming that [tex]g = 9.81\; \rm m \cdot s^{-2}[/tex].)

Explanation:

Let the initial velocity of this stone be [tex]v\; \rm m \cdot s^{-1}[/tex] ([tex]v \ge 0[/tex] because speed should be non-negative.) That should be numerically equal to the initial horizontal speed of this stone. Because the stone was thrown horizontally, its vertical speed would initially be zero ([tex]0\; \rm m \cdot s^{-1}[/tex].)

The weight of this stone will accelerate the stone downwards. Because the weight of the stone is in the vertical direction, it will have no effect on the horizontal speed of the stone.

Therefore, if air resistance is indeed negligible, the horizontal speed of this stone will stay the same. The horizontal speed of the stone after [tex]0.5\; \rm s[/tex] should still be [tex]v\; \rm m \cdot s^{-1}[/tex]- same as its initial value.

On the other hand, if [tex]g = 9.81\; \rm m \cdot s^{-2}[/tex], the (downward) vertical speed of this stone will increase by [tex]9.81\; \rm m \cdot s^{-2}[/tex] every second. After [tex]0.5\; \rm s[/tex], the vertical speed of this stone would have become:

[tex]0.5\; \rm s \times 9.81\; \rm m \cdot s^{-2} = 4.905\; \rm m \cdot s^{-1}[/tex].

[tex]t = 0\; \rm s[/tex]:

[tex]t = 0.5\; \rm s[/tex]:

Refer to the diagram attached. Consider the vertical speed and the horizontal speed of this rock as lengths of the two legs of a right triangle. The numerical value of the velocity of this stone would correspond to the length of the hypotenuse of that right triangle. Apply the Pythagorean Theorem to find that numerical value:

[tex]\begin{aligned} &\text{numerical value of velocity} \\ &= \sqrt{{\left(\text{horizontal speed}\right)}^2 + {\left(\text{vertical speed}\right)}^2}\end{aligned}[/tex].

At [tex]t = 0\; \rm s[/tex], the magnitude of the velocity of this stone would be [tex]\sqrt{v^{2} + 0^2} = v\; \rm m \cdot s^{-1}[/tex].

At [tex]t = 0.5\; \rm s[/tex], the magnitude of the velocity of this stone would be [tex]\left(\sqrt{v^2 + 4.905^2}\right)\; \rm m \cdot s^{-1}[/tex].

The question requires that the magnitude of the velocity of the stone at [tex]t = 0.5\; \rm s[/tex] should be [tex]1.5[/tex] times the value at [tex]t = 0\; \rm s[/tex]. In other words:

[tex]\sqrt{v^2 + 4.905^2} = 1.5 \, v[/tex].

Square both sides of this equation and solve for [tex]v[/tex]:

[tex]v^2 + 4.905^2 = 2.25\, v^2[/tex].

[tex]v \approx 4.39[/tex] ([tex]v \ge 0[/tex] given that speed should be non-negative.)

Therefore, the initial velocity of this stone should be approximately [tex]4.39\; \rm m \cdot s^{-1}[/tex].

Answer:

Workdone = 25500Nm

Explanation:

Given the following data

Force = 425N

Distance = 60m

To find the workdone

Workdone = force *distance

Substituting into the equation, we have

Workdone = 425*60

Workdone = 25500Nm

Answer:

people who had a hard time to help other people because they feel that if they help

that the work of the person is much harder than his own work

Individuals who find it difficult to help others because they believe that if they do, the person's job will be much tougher than their own.

A group of people working together to achieve a shared purpose.

Collaboration, cooperation, and coordination are all words that come to mind when thinking of teaming.

Learn more:

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In,radioactive decay,the half-life is the length of time after which there is a 50 percent chance that an atom will have undergone nuclear decay.It varies depending on the atom type and isotope,and is usually determined experimentally.

Answer:

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

Explanation:

Under the assumption that no external forces are exerted on both the small object and the big object, whose situation is described by the Principle of Momentum Conservation:

[tex]p_{S,1}+p_{B,1} = p_{S,2}+p_{B,2}[/tex] (1)

Where:

[tex]p_{S,1}[/tex], [tex]p_{S,2}[/tex] - Initial and final momemtums of the small object, measured in kilogram-meters per second.

[tex]p_{B,1}[/tex], [tex]p_{B,2}[/tex] - Initial and final momentums of the big object, measured in kilogram-meters per second.

If we know that [tex]p_{S,1} = 7\,\frac{kg\cdot m}{s}[/tex], [tex]p_{B,1} = 0\,\frac{kg\cdot m}{s}[/tex] and [tex]p_{S, 2} = 4\,\frac{kg\cdot m}{s}[/tex], then the final momentum of the big object is:

[tex]7\,\frac{kg\cdot m}{s} + 0\,\frac{kg\cdot m}{s} = 4\,\frac{kg\cdot m}{s}+p_{B,2}[/tex]

[tex]p_{B,2} = 3\,\frac{kg\cdot m}{s}[/tex]

The magnitude of the large object's momentum change is:

[tex]p_{B,2}-p_{B,1} = 3\,\frac{kg\cdot m}{s}-0\,\frac{kg\cdot m}{s}[/tex]

[tex]p_{B,2}-p_{B,1} = 3\,\frac{kg\cdot m}{s}[/tex]

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

Answer:

τ = (7.96 x 10⁴ m⁻³)T

This is the expression for maximum allowable shear stress in terms of the maximum torque applied in Nm.

Explanation:

The maximum allowable shear stress on the solid shaft can be given by the torsional formula as follows:

τ = Tc/J

where,

τ = Maximum Allowable Shear Stress = ?

T = Maximum Torque Applied to the Shaft

c = maximum distance from center to edge = radius in this case = 20 mm = 0.02 m

J = Polar Moment of inertia = πr⁴/2 = π(0.02 m)⁴/2 = 2.51 x 10⁻⁷ m⁴

Therefore,

τ = T(0.02 m)/(2.51 x 10⁻⁷ m⁴)

τ = (7.96 x 10⁴ m⁻³)T

This is the expression for maximum allowable shear stress in terms of the maximum torque applied in Nm.

Answer:

Globel warming

Explanation:

hope this helpex