You are pushing on a heavy door, trying to slide it open. If your friend stands behind you and helps you push, how have the forces changed?
Question options:
There will be less friction.
The applied force does not change.
The net force will decrease.
The net force applied will increase.
Answer: Option (d) is the correct answer.
Explanation:
Net force is defined as the sum of total number of forces acting on a substance.
For example, when you and your friend are pushing a door in order to slide it open then it means that the net force is sum of force applied by you and force applied by your friend.
Therefore, the force applied earlier was less but it increase when your friend also started applying force.
Thus, we can conclude that if your friend stands behind you and helps you push, then the net force applied will increase.
A car is traveling with a constant speed when the driver suddenly applies the brakes, causing the car to slow down with a constant acceleration of magnitude 3.50 m/s2. if the car comes to a stop in a distance of 30.0 m, what was the car's original speed?
The car's original speed can be calculated by utilizing the kinematic equation which is a principle in physics. Given that the car came to a stop with constant deceleration in a certain distance, the estimated initial speed was approximately 26.9 m/s.
Explanation:The subject of the question involves concepts in physics specifically related to the equations of motion. To solve this, we can utilize the kinematic equation which relates initial velocity, acceleration, and distance: v2 = u2 + 2as, where 'v' is the final velocity, 'u' is the initial velocity, 'a' is acceleration and 's' is distance. Since the car comes to a stop, our final velocity v = 0 m/s. We know that the distance s = 30m and the acceleration is given as a = -3.50 m/s2 (negative because it's deceleration). From these, we can find the initial velocity 'u'. When we substitute these values into the equation, we can solve for 'u' as √(v2 - 2as), which equals approximately √(0 - 2*-3.50*30) ≈ 26.9 m/s. Therefore, the car's original speed was approximately 26.9 m/s.
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What is the energy of a photon with the wavelength 150 nm?
At a given point on a horizontal streamline in flowing air, the static pressure is â2.0 psi (i.e., a vacuum) and the velocity is 150 ft/s. determine the pressure at a stagnation
The pressure at the stagnation point is 149.854 lb/ft².
Explanation:According to Bernoulli's equation, the total pressure at a given point is the sum of the static pressure and the dynamic pressure. The dynamic pressure is given by 0.5 * ρ * v^2, where ρ is the density of the fluid and v is the velocity of the fluid. In this case, the static pressure is -2.0 psi (a vacuum) and the velocity is 150 ft/s. The density of air at sea level is approximately 1.14 kg/m³, which is equivalent to 0.075 lb/ft³. Converting the units, we have -2.0 psi = -18.896 lb/ft². Plugging in the values into Bernoulli's equation, we can calculate the dynamic pressure:
Dynamic pressure = 0.5 * (0.075 lb/ft³) * (150 ft/s)^2 = 168.75 lb/ft².
The total pressure at the stagnation point is equal to the sum of the static pressure and the dynamic pressure. Therefore, the pressure at the stagnation point is:
Total pressure = -18.896 lb/ft² + 168.75 lb/ft² = 149.854 lb/ft².
A vector A is added to B=6i-8j. The resultant vector is in the positive x direction and has a magnitude equal to A . What is the magnitude of A?
a)11
b)7.1
c)5.1
d)8.3
e)12.2
The correct answer would be letter d, 8.3.
Solution for the problem follows:
Given are:
B = 6i - 8j
A is unknown; let A be = mi + nj
A+B is along the x axis (therefore A+B = Ki + 0j, where K is unknown,
but then again the magnitude of A+B is the similar as the magnitude of A,
so mag(A+B)=K=sqrt(m^2+n^2), or K^2 = m^2+n^2.
A+B, from simple vector addition, will be now (m+6)i + (n-8)j.
Ever since we previously
know A+B = Ki + 0j, we now know that:
m+6 = K
n-8 = 0, which implies n=8.
Thus, K^2=m^2+n^2 ====> (m+6)^2 = m^2 +8^2
= m^2 + 12m + 36 = m^2 + 64
= 12m = 28
= m = 2.33333...
Therefore, the magnitude of A is sqrt[(2.333...)^2 + 8^2] = 8.3333
The magnitude of [tex]\vec A[/tex] is [tex]8.3[/tex].
Further explanation:
A vector is a quantity having magnitude and direction both. It is represented as the product of magnitude and direction vector.
Given:
The vector is given as [tex]\vec B = 6\hat i - 8\hat j[/tex].
The direction of resultant vector is in the X- direction.
Concept used:
Consider a vector [tex]\vec A = a\hat i + b\hat j[/tex] which is added to [tex]\vec B[/tex] to get a resultant vector [tex]\vec C[/tex]. The resultant vector is directed in the positive X-direction which means that the y- component of the vector is zero.
The expression for the resultant vector is given as.
[tex]\vec C = \vec A + \vec B[/tex]
Substitute [tex]6\hat i - 8\hat j[/tex] for [tex]\vec B[/tex] in the above expression.
[tex]\begin{gathered}\vec C = \left( {a\hat i + b\hat j} \right) + \left( {6\hat i - 8\hat j} \right) \\= \left( {a + 6} \right)\hat i + \left( {b - 8} \right)\hat j \\ \end{gathered}[/tex]
The resultant vector is represented as.
[tex]\vec C = x\hat i + 0\hat j[/tex]
Compare the above two expression of resultant vector.
[tex]x = a + 6[/tex] …… (1)
[tex]\begin{aligned}0&=b-8\\b&=8\\\end{aligned}[/tex]
The magnitude of resultant vector is equal to the magnitude of [tex]\vec A[/tex].
The expression for the magnitude of [tex]\vec A[/tex] is given as.
[tex]\left| {\vec A} \right| = \sqrt {\left( {{a^2}} \right) + \left( {{b^2}} \right)}[/tex] …… (2)
The expression for the magnitude of [tex]\vec C[/tex] is given as.
[tex]\left| {\vec C} \right| = \sqrt {{{\left( {a + 6} \right)}^2}}[/tex]
Compare the above two expressions.
[tex]\sqrt {\left( {{a^2}} \right) + \left( {{b^2}} \right)}= \sqrt {{{\left( {a + 6} \right)}^2}}[/tex]
Substitute 8 for [tex]b[/tex] in above expression.
[tex]\begin{aligned}\sqrt {\left( {{a^2}} \right) + \left( {{8^2}} \right)}&=\sqrt {{{\left( {a + 6} \right)}^2}}\\{a^2} + 64&={a^2} + 36 + 12a \\12a&=28 \\a&=2.33 \\ \end{aligned}[/tex]
Substitute 2.33 for a and 8 for b in equation (2).
[tex]\begin{aligned}\left| {\vec A} \right|&=\sqrt {{{\left( {2.33} \right)}^2} + {{\left( 8 \right)}^2}}\\&=8.33 \\ \end{aligned}[/tex]
Thus, the magnitude of vector A is [tex]8.33[/tex].
Learn more:
1. Motion under friction https://brainly.com/question/7031524.
2. Conservation of momentum https://brainly.com/question/9484203.
3. Circular motion https://brainly.com/question/9575487.
Answer Details:
Grade: College
Subject: Physics
Chapter: Vectors
Keywords:
Vectors, product, magnitude, direction, resultant vector, adding vector, subtraction of vector, 2.33, 8, 8.33, 8.3, 8.33.
A 30-kg chandelier hangs from a ceiling on a vertical 4.4-m-long wire. what horizontal force would be necessary to displace its position 0.16 m to one side?
The horizontal force would be necessary to displace its position 0.16 m to one side will be 10.7 N.
What is horizontal force?Horizontal force is defined as a force applied in a direction parallel to the horizon. A force exerted on a body has two components: a horizontal component and a vertical component.
Given that,
m = 30 kg,
W = mg = (30 kg)*(9.8 m/s²)
W = 294 N, the weight of the chandelier.
Let T = tension in the wire.
Let F = the horizontal force required to displace the chandelier by 0.16 m.
The angle that the wire makes with the vertical is given by:
sin(θ) = 0.16/4.4 = 0.0364
θ = sin⁻¹ 0.0364
θ = 2.084°
For equilibrium,
T cos(θ) = W
0.9993T = 294
T = 294.195 N
We know that,
F = Tsin(θ)
F = 294.195*0.0364
F = 10.709 N
Thus, The horizontal force would be necessary to displace its position 0.16 m to one side will be 10.7 N.
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When opening a door, you push on it perpendicularly with a force of 57.0 n at a distance of 0.470 m from the hinges. what torque (in n·m) are you exerting relative to the hinges? (enter the magnitude.)?
What is the direction of the centripetal force when applied to an object?
the direction opposite to the object’s velocity
the same direction as the object’s velocity
perpendicular to an object’s motion
parallel with an object’s motion
The direction of the centripetal force when applied to an object is perpendicular to the object's motion.
What is centripetal force?Any motion along a curved road is accelerated, necessitating the application of force to the path's center of curvature. This force is known as the centripetal force, which is a force that seeks its center.
The direction of centripetal force, or even net force, is always that of acceleration. As an illustration, consider how quickly the earth is moving toward the sun. We are not moving toward the sun since the acceleration vector for planet Earth is constantly pointing in that direction and the velocity is perpendicular to it.
When force applied to an item, the centripetal force's direction is perpendicular to the object's motion.
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During an episode of turbulence in an airplane you feel 170 n heavier than usual. if your mass is 73 kg, what are the magnitude and direction of the airplane's acceleration?
Final answer:
The airplane's acceleration that caused the passenger to feel 170 N heavier is 2.33 m/s² upwards.
Explanation:
When a person experiences an increase in weight due to airplane turbulence, it indicates that there's an additional force acting on them because of the airplane's acceleration. To find the magnitude and direction of the airplane's acceleration, we can use Newton's second law of motion (F=ma), where F is the force, m is the mass, and a is the acceleration.
The student feels 170 N heavier, implying that the upward acceleration of the airplane is causing this additional force. Since the student's mass is 73 kg, we can calculate the acceleration using the formula a = F/m, where F is the additional force (170 N) and m is the mass of the student (73 kg). Hence, a = 170 N / 73 kg = 2.33 m/s². This is the magnitude of the acceleration.
The direction of the acceleration is upwards, as it increases the normal force which is felt as an increase in apparent weight.
Calculate the efficiency of an engine in a power plant operating between 40 ° c and 320 °
c. remember that the efficiency is a decimal number less than 1.
A 1.60-m-long barbell has a 25.0 kg weight on its left end and a 37.0 kg weight on its right end. you may want to review ( pages 204 - 205) . part a if you ignore the weight of the bar itself, how far from the left end of the barbell is the center of gravity
The solid aluminum shaft has a diameter of 50mm and an allowable shear stress of 60 mpa. determine the largest torque t1 that can be applied to the shaft if it is also subjected to the other torsional loadings
Final answer:
The largest torque that can be applied to the solid aluminum shaft is 1.5 MPa.
Explanation:
To determine the largest torque that can be applied to the solid aluminum shaft, we need to consider the allowable shear stress and the diameter of the shaft. The formula to calculate torque is t = rF sin 0, where t is the torque, r is the radius, F is the applied force, and 0 is the angle between the force and the radius. In this case, since the force is perpendicular to the radius, the angle is 90 degrees, so sin 0 = 1.
The diameter of the shaft is given as 50mm, which means the radius is 25mm. We need to convert the radius to meters, so the radius is 25/1000 = 0.025m. The allowable shear stress is given as 60MPa.
Using the formula for torque, t = rF = (0.025m)(60MPa) = 1.5MPa.
Therefore, the largest torque that can be applied to the shaft is 1.5MPa.
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A tuba makes a low pitched sound and a flute makes a high pitched sound. Which statement accurately describes the wavelengths of the two sounds?
A) The wavelengths of the two sounds are the same.
B) The wavelength of the high pitched sound is not measureable.
C) The wavelength of the low pitched sound is longer than the wavelength of the high pitched sound.
D) The wavelength of the low pitched sound is shorter than the wavelength of the high pitched sound.
A car slams on its brakes, coming to a complete stop in 4.0 s. The car was traveling south at 60.0 mph. Calculate the acceleration.
Answer:
Acceleration, [tex]a=-6.705\ m/s^2[/tex]
Explanation:
It is given that,
Velocity of the car, u = 60 mph = 26.82 m/s
Finally it comes to stop, v = 0
Time taken, t = 4 s
We need to find the acceleration of the car. The change in velocity divided by time is called the acceleration of the object. Mathematically, it is given by :
[tex]a=\dfrac{v-u}{t}[/tex]
[tex]a=\dfrac{0-26.82}{4}[/tex]
[tex]a=-6.705\ m/s^2[/tex]
Negative sign shows the car is decelerating. So, the acceleration of the car is [tex]6.705\ m/s^2[/tex]. Hence, this is the required solution.
Two friends, barbara and neil, are out rollerblading. with respect to the ground, barbara is skating due south at a speed of 6.3 m/s. neil is in front of her. with respect to the ground, neil is skating due west at a speed of 2.4 m/s. find neil's velocity ((a) magnitude and (b) direction relative to due west, as seen by barbara.
Final answer:
To find Neil's velocity relative to Barbara, their individual velocities are considered as vectors and subtracted. Neil's relative velocity is calculated to be approximately 6.74 m/s at 69° north of west.
Explanation:
To determine Neil's velocity relative to Barbara, we need to consider both Neil’s and Barbara's velocities as vectors and subtract the velocity of Barbara from Neil's velocity. Barbara is moving due south at 6.3 m/s, and Neil is moving due west at 2.4 m/s. By defining south as the negative y-direction and west as the negative x-direction, we can represent Barbara’s velocity vector as (0 m/s, -6.3 m/s) and Neil's as (-2.4 m/s, 0 m/s).
To find Neil’s velocity relative to Barbara, we calculate the vector difference: Neil’s velocity minus Barbara's velocity:
VNB = VN - VB = (-2.4 m/s, 0 m/s) - (0 m/s, -6.3 m/s) = (-2.4 m/s, 6.3 m/s)
To find the magnitude of Neil's relative velocity (VNB), we use the Pythagorean theorem: |VNB| = √((-2.4 [tex]m/s^2[/tex] + (6.3 [tex]m/s^2[/tex]), which gives:
|VNB| = √(5.76 + 39.69) = √(45.45) ≈ 6.74 m/s
The direction relative to due west can be found using the arctangent of the y-component over the x-component:
θ = arctan(6.3 / 2.4) ≈ arctan(2.625) ≈ 69°
Therefore, Neil's velocity relative to Barbara is approximately 6.74 m/s at an angle of 69° north of west.
A 2kg particle moving along the x-axis experiences the net force shown at right. The particle’s velocity is 3.0m/s at x = 0m.
A) At what position is the particle moving the fastest?
B) What is the particle’s velocity at x=8m? Show your work