Explanation:
We have,
Mass of a baseball is 0.147 kg
Initial velocity of the baseball is 44.5 m/s
The ball is moved in the opposite direction with a velocity of 55.5 m/s
It is required to find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.
Change in momentum,
[tex]\Delta p=mv-mu\\\\\Delta p=m(v-u)\\\\\Delta p=0.147\times ((-55.5)-44.5)\\\\\Delta p=-14.7\ kg-m/s\\\\|\Delta p|=14.7\ kg-m/s[/tex]
Impulse = 14.7 kg-m/s
Therefore, the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat is 14.7 kg-m/s
Students who participated in an optimism workshop:
O
A. had fewer episodes of depression.
O
B. had fewer health problems.
O
C. All of these
O
D. had fewer episodes of anxiety.
Answer:
C
Explanation:
Researches have shown that Optimism may lead to the advancement of a balanced lifestyle and proactive behavior, cognitive responses combined with improved resilience and problem solving abilities and a more successful processing of negative input will make a huge difference to mental and physical well-being.
So, Students who participated in an optimism workshop had fewer episodes of depression, fewer health problems and had fewer episodes of anxiety.So, the correct answer is C.
The answer is All Of These
compare magnetic properties of iron and steel
Answer: Iron and steel are two separate metallic elements.
Explanation:
The following are the differences between magnetic properties of iron and steel.:
1. Iron get magnitized quickly under the influence of magnetic field whereas the steel magnitized slowly.
2. The iron looses its magnitism when magnetic field is removed but steel retains the magnitism for long when magnetic field is removed.
The force exerted by the wind on the sails of a sailboat is Fsail = 330 N north. The water exerts a force of Fkeel = 210 N east. If the boat (including its crew) has a mass of 260 kg, what are the magnitude and direction of its acceleration? magnitude m/s2 direction ° north of east
Answer:
The magnitude of the acceleration is [tex]a_r = 1.50 \ m/s^2[/tex]
The direction is [tex]\theta = 32.5 6^o[/tex] north of east
Explanation:
From the question we are told that
The force exerted by the wind is [tex]F_{sail} = (330 ) \ N \ north[/tex]
The force exerted by water is [tex]F_{keel} = (210 ) \ N \ east[/tex]
The mass of the boat(+ crew) is [tex]m_b = 260 \ kg[/tex]
Now Force is mathematically represented as
[tex]F = ma[/tex]
Now the acceleration towards the north is mathematically represented as
[tex]a_n = \frac{F_{sail}}{m_b}[/tex]
substituting values
[tex]a_n = \frac{330 }{260}[/tex]
[tex]a_n = 1.269 \ m/s^2[/tex]
Now the acceleration towards the east is mathematically represented as
[tex]a_e = \frac{F_{keel}}{m_b }[/tex]
substituting values
[tex]a_e = \frac{210}{260}[/tex]
[tex]a_e =0.808 \ m/s^2[/tex]
The resultant acceleration is
[tex]a_r = \sqrt{a_e^2 + a_n^2}[/tex]
substituting values
[tex]a_r = \sqrt{(0.808)^2 + (1.269)^2}[/tex]
[tex]a_r = 1.50 \ m/s^2[/tex]
The direction with reference from the north is evaluated as
Apply SOHCAHTOA
[tex]tan \theta = \frac{a_e}{a_n}[/tex]
[tex]\theta = tan ^{-1} [\frac{a_e}{a_n } ][/tex]
substituting values
[tex]\theta = tan ^{-1} [\frac{0.808}{1.269 } ][/tex]
[tex]\theta = tan ^{-1} [0.636 ][/tex]
[tex]\theta = 32.5 6^o[/tex]
A bullet of 5 gm is fired from a pistol of 1.5 kg. If the recoil velocity of pistol is 1.5 m/s,
find the velocity of bullet.
Answer:
-450 m/s
Explanation:
Momentum is conserved.
p₀ = p
0 = (1.5 kg) (1.5 m/s) + (0.005 kg) v
v = -450 m/s
Which explains what a velocity time graph would look like with no acceleration?
O curved line on a upward slope
O straight line on a downward slope
O straight line with no slope
O curved line on a downward slope
Answer:
Hey!
Your answer is C!
Explanation:
HOPE THIS HELPS!!
Straight line with no slope explains what a velocity time graph would look like with no acceleration. The correct option is 3.
When there is no acceleration, the item is travelling at a steady speed. The y-axis of a velocity-time graph represents velocity, while the x-axis represents time.
A straight line with no slope implies constant velocity, where the velocity remains constant over time. There is no change in velocity, which indicates that there is no acceleration.
Thus, on the velocity-time graph, a straight line with no slope depicts an object travelling with no acceleration. The correct option is 3.
For more details regarding velocity-time graph, visit:
https://brainly.com/question/33512397
#SPJ2
Your question seems incomplete, the probable complete question is:
Which explains what a velocity time graph would look like with no acceleration?
O curved line on a upward slopeO straight line on a downward slopeO straight line with no slopeO curved line on a downward slopeA student, starting from rest, slides down a water slide. On the way down, a kinetic frictional force (a nonconservative force) acts on her. The student has a mass of 74.0 kg, and the height of the water slide is 11.3 m. If the kinetic frictional force does -5.42 × 103 J of work, how fast is the student going at the bottom of the slide?
Answer:
The student is going at the bottom of the slide with a velocity of 8.66 m/s
Explanation:
Given;
mass of the student, m = 74 kg
height of the water slide, h = 11.3 m
work done, W = -5.42 × 10³ J
Apply work energy theorem;
[tex]W = \frac{1}{2}mv_f^2+ mgh_f-(\frac{1}{2}mv_o^2 + mgh_o)\\\\ W + \frac{1}{2}mv_o^2 + mgh_o -mgh_f= \frac{1}{2}mv_f^2\\\\ W + \frac{1}{2}mv_o^2 +mg(h_o-h_f) = \frac{1}{2}mv_f^2\\\\\frac{W}{m} + \frac{1}{2} v_o^2 + g(h_o-h_f) = \frac{1}{2}v_f^2\\\\\frac{2W}{m}+ v_o^2 + 2g(h_o-h_f) = v_f^2\\\\v_f = \sqrt{\frac{2W}{m}+ v_o^2 - 2g(h_f-h_o)} \\\\v_f = \sqrt{\frac{2(-5.42*10^3)}{74}+ (0)^2 - 2*9.8(-11.3)}\\\\v_f=\sqrt{-146.4865+221.48} \\\\v_f = \sqrt{74.9935} \\\\v_f = 8.66 \ m/s[/tex]
Therefore, the student is going at the bottom of the slide with a velocity of 8.66 m/s
Two dogs play tug of war with a rope toy that has a mass of 0.50 kg. If one dog pulls on the toy with a force of 140.0 N, and the other dog pulls in the opposite direction with a force of 138.0 N, what is the horizontal acceleration of the toy?
Answer:[tex]4\ m/s^2[/tex]
Explanation:
Given
mass of toy [tex]m=0.5\ kg[/tex]
Force by first dog [tex]F_1=140\ N[/tex]
Force by second dog is [tex]F_2=138\ N[/tex]
Net force on the dog is
[tex]F_{net}=F_1-F_2=140-138[/tex]
[tex]F_{net}=2\ N[/tex]
and [tex]F_{net}=ma[/tex]
[tex]a=\dfrac{F_{net}}{m}[/tex]
[tex]a=\dfrac{2}{0.5}[/tex]
[tex]a=4\ m/s^2[/tex]
So, horizontal acceleration of the toy is [tex]4\ m/s^2[/tex]
Particle A of charge 3.30 10-4 C is at the origin, particle B of charge -6.24 10-4 C is at (4.00 m, 0), and particle C of charge 1.06 10-4 C is at (0, 3.00 m). We wish to find the net electric force on C. (a) What is the x component of the electric force exerted by A on C?
Answer:
a) [tex]E_T=134,484\frac{N}{C}\hat{i}+149954.66\frac{N}{C}\hat{j}[/tex]
b) zero
Explanation:
a) To find the electric field at point C, you sum the contribution of the electric fields generated by the other two charges. The total electric field at C is given by:
[tex]E_T=E_1+E_2[/tex]
E1: electric field of charge 1
E2: electric field of charge 2
It is necessary to calculate the x and y components of both E1 and E2. You take into account the direction of the fields based on the charge q1 and q2:
[tex]E_1=k\frac{q_1}{r_{1,3}}[cos\theta\hat{i}+sin\theta \hat{j}]\\\\E_2=k\frac{q_2}{r_{2,3}}[cos\phi\hat{i}-sin\phi \hat{j}]\\\\[/tex]
r13: distance between charges 1 and 3
r12: charge between charges 2 and 3
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
Thus, you first calculate the distance r13 and r23, and also the angles:
[tex]r_{1,3}=3.00m\\\\r_{2,3}=\sqrt{(3.00m)^2+(4.00m)^2}=5.00m\\\\\theta=90\°\\\\\phi=tan^{-1}(\frac{4.00}{3.00})=53.13\°[/tex]
Next, you replace the values of all parameters in order to calculate E1 and E2:
[tex]E_1=(8.98*10^9Nm^2/C^2)(\frac{3.30*10^{-4}C}{(3.00m)^2})\hat{j}\\\\E_1=329266.66\frac{N}{C}\\\\E_2=(8.98*10^9Nm^2/C^2)(\frac{6.24*10^{-4}C}{(5.00m)^2})[cos53.13\°\hat{i}-sin(53.13\°)\hat{j}]\\\\E_2=224140.8[0.6\hat{i}-0.8\hat{j}]=134484\hat{i}-179312\hat{j}[/tex]
finally, you obtain for ET:
[tex]E_T=134,484\frac{N}{C}\hat{i}+(329266.66-179312)\frac{N}{C}\hat{j}\\\\E_T=134,484\frac{N}{C}\hat{i}+149954.66\frac{N}{C}\hat{j}[/tex]
b) The x component of the force exerted by A on C is zero because there is only a vertial distance between them. Thus, there is only a y component force.
2.A 5 Ohm resistor is connected to a 9 Volt battery. How many Joules of thermal energy are produced in 7 minutes?
3.The current in a flashlight powered by 4.5 Volts is 0.5 A. What is the power delivered to the flashlight?
4.If the flashlight in the previous problem is left on for 3 minutes, how much electric energy is delivered to the bulb?
7.A 50 Ohm resistance causes a current of 5 milliamps to flow through a circuit connected to a battery. What is the power in the circuit?
8.How many Joules of electric energy are delivered to a 60 Watt lightbulb if the bulb is left on for 2.5 hours?
Answer:
Explanation:
2 )
power of an electric device = V² / R where V is volts and R is resistance
putting given data
power = 9²/ 5
= 16.2 J/s
energy produced in 7 minutes
= 16.2 x 7 x 60
= 6804J .
3 ) Power of an electrical device
= V² / R
= V X I where I is current
= 4.5 x .5
= 2.25 W or J/s
4 )
energy used in 3 minutes with power of 2.25 W
= 2.25 x 3 x 60
= 405 J .
7 )
power of a electrical device
= V x I
IR x I where R is resistance .
= I²R
putting given data
power = .005² x 50
= 1.25 x 10⁻³ W .
8 )
Energy used up by a 60 W bulb in 2.5 hours
= 60 x 2.5 x 60 x 60
= 5.4 x 10⁵ J .
Coulomb's law: A point charge Q is located a short distance from a point charge 3 Q, and no other charges are present. If the electrical force on Q is F, what is the electrical force on 3 Q
Answer:
The electrical force on 2nd charge (3 Q) will also be equal to F.
Explanation:
The electrical force between two electrically charged particles is given by the Coulomb's Law. The mathematical formula of Coulomb's Law is as follows:
F = kQ₁Q₂/r²
where,
F = Electrical Force exerted by both particles on each other
k = Coulomb's Constant
Q₁ = Magnitude of 1st Charge = Q
Q₂ = Magnitude of 2nd Charge = 3 Q
r = Distance between the charges
Therefore,
F = k(Q)(3 Q)/r²
F = 3 kQ²/r²
Since, it is clear from the formula that the magnitude of force applied by the first charge on 2nd charge is equal to the force applied by 2nd charge on 1st charge.
Therefore, the electrical force on 2nd charge (3 Q) will also be equal to F.
With the wire back at its initial location, you connect a second identical battery in series with the first one. When you close the switch, how does the new angle of deflection of the north pole of the compass needle compare to its initial deflection
Answer:
It is larger
Explanation:
I believe the answer would be "It is larger" since v = 2v (since it's in series), which would increase the magnetic field and thus produced a bigger angle.
If a spaceship has a momentum of 30,000 kg-m/s to the right and a mass of
400 kg, what is the magnitude of its velocity?
A. 12,000,000 kg-m/s
B. 75 m/s
C. 1,200 kg-m/s
D. 300 m/s
Answer:
75m/s
Explanation:
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