Solve 100N up 90N down for net force

The force on the driver is 7629 N. The velocity of the second ball is 3.6 m/s. The combined velocity of the balls is 13.37 m/s.

We have to find the acceleration using;

v = u - at

v = final velocity = 0 m/s

u = initial velocity = 35.6m/s

a = acceleration = ?

t = time = 0.35 s

u = at

a = u/t = 35.6m/s / 0.35 s

a = 101.7 ms-2

The force on the driver =  75kg ×  101.7 ms-2 = 7629 N

Using the principle of conservation of momentum;

Momentum before collision = momentum after collision

m1u1 +m2u2 = m1v1 + m2v2

Hence

(3.2 × 4.1) + 0 = (3.2 × 1.5) + 2.3v2

13.12 = 4.8 + 2.3v2

13.12 - 4.8 = 2.3v2

v2 = 13.12 - 4.8/2.3

v2 = 3.6 m/s

Using the principle of conservation of linear momentum;

m1u1 + m2u2 = m1v1 + m2v2

(4.3 × 18.6) + (5.8 × 9.5) = (4.3 + 5.8) v

v = 79.98 + 55.1/10.1

v = 13.37 m/s

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The answer is D

I took the quiz lol

Answer:

Answer below

Explanation:

We can rearrange this statement to get the angular velocity equation: = r v / |r|2, where all of these variables are vectors and |r| is the radius's absolute value. The angular velocity is really a pseudovector with a direction perpendicular to the plane of rotational movement.

The rotating movement of bodies is described by angular velocity. It determines how quickly they rotate around a central point. Rotations may be divided into two categories. The first explains the motion of an object's centre of mass around a certain point in space, which is referred to as an origin. Planets orbiting the Sun, for example, or an automobile exiting a roadway are two instances.

The first angular velocity equation is similar to the linear velocity equation: = (2 - 1) / t = / t, where 1 and 2 are two values of angles on a circle, and is the difference between them. t is the time it takes for the angle to shift. As you can see, there is a ratio of the positional change in a period for normal velocity, but we utilise angle instead of distance here. Definition of angular velocity using linear velocity The second angular velocity formula, v = r, may be obtained from the linear velocity and radius relationship using the cross product.

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(Hope this helps can I pls have brainlist (crown)☺️)

Answer:

The farther out a galaxy is from Earth, the faster it is moving away.

Explanation:

Answer:

100 m/s

Explanation:

Speed increases at a rate of 10 m/s (actually 9.8 m/s) every second. Thus after 10 seconds, the speed is 10 x 10 = 100 m/s.

The first resonant position below the open end of the resonance tube is; one-quarter of the wavelength

In the event of the first resonant position in a resonance tube, there will be a maximum air displacement which is only one antinode right at the open end where the motion is constrained.

However, there will be no displacement at the closed end which means another one node right at the closed end where air is halted.

This means that the standing wave will have one-quarter of the wavelength in the test tube.

Thus;

L = ¼λ

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Hi there!

We can use impulse for this situation:

I = Δp = mΔv

Impulse = Force × time, so:

I = 63.9(24) = 1533.6 Ns

Find force by dividing by time:

I/t = 1533.6/1.2 = 1278 N

Answer:

H = V t - 1/2 g t^2         V and g are in different directions

-8.2 = V * 6.2 - 1/2 * 10 * 6.2*2 = 6.2 V - 192.2

V = 184 / 6.2 = 29.7 m/s

Check: find time for ball to return to zero height

0 = 29.7 T - 5 T^2

T = 29.7 / 5 = 5.94 sec

The ball must have fallen 8.2 m in (6.2 - 5.94) sec = .26 sec

S = 29.7 * .26 + 5 * .26^2 = 8.1

Answer:

it's ans is D because all three are wrong

The work done on the given gas in one cycle is -405.3 kJ.

The given parameters;

Convert the pressure to Pascal (N/m²);

1 atm = 101325

The work done in one cycle is the area of the triangle and it is calculated as follows

[tex]Area = \frac{1}{2} \times base \times height\\\\Area = \frac{1}{2} \times (3 \ m^3\ -\ 1 \ m^3)\times (6 \ atm \ - \ 2 \ atm)\\\\Area = 4 \ atm -m^3\\\\Area = 4 \ atm -m^3 \ \times \frac{101325 \ Pa}{1 \ atm} \\\\Area = 405,300 \ m^3.Pa\\\\Area = 405,300 \ m^3. (N/m^2)\\\\Area = 405,300 \ Nm\\\\Area = 405,300 \ J\\\\Area = 405.3 \ kJ[/tex]

the net work done on the gas = - 405.3 kJ

Thus, the work done on the given gas in one cycle is -405.3 kJ.

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The acceleration of both the rocks is only due to the gravitational force acting on them. Even though the speed of thrown rock would be greater than that of dropped rock, the acceleration of both the rocks would be g downwards.

Answer:

1.80

Explanation:

38.4=-12.5(a)-1/2(-9.80)(a)^2

a= acceleration
^2 is the exponent

Answer:

c

Explanation:

Answer: The area of the edge of a knife's blade is extremely small.  

Syringes are used to take blood for blood tests.  

When air is sucked out of a drinking straw, the air pressure inside if decreases and the atmospheric pressure outside forces the liquid to go inside the straw.

Explanation:

The water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.

The given parameters;

The vertical pressure of the water is calculated as follows;

[tex]P = \rho gh\\\\\frac{P}{h} = \rho g\\\\\frac{P}{h} = k\\\\\frac{P_1}{h_1} = \frac{P_2}{h_2} \\\\[/tex]

The vertical height of the first floor from the 13th floor = 130 ft

The vertical height of the 13 ft floor = 10  ft

[tex]P_1 = \frac{P_2 h_1}{h_2} \\\\P_1 = \frac{35 \times 130}{10} \\\\P_1 = 455 \ PSI[/tex]

Thus, the water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.

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Answer:

The equation net τ=ΔLΔt net τ = Δ L Δ t gives the relationship between torque and the angular momentum produced.

In trajectory, acceleration vector is not the same with the velocity vector but it depends on the change in velocity. The acceleration path is perpendicular to the velocity vector path. The Correct answers are:

a.) D

b.) B

c.) C

d.) A

A Vector is a quantity that has both magnitude and direction.

Given that a  ball is fired from a cannon at point 1 and follows the trajectory shown in the figure.

a.) The vector best represents the ball's velocity at position 2 is D

b.) The vector best represents the ball's acceleration at point 2 is B

c.) Which vector best represents the ball's velocity at position 3 = C

d.) Which vector best represents the ball's acceleration at point  = A

In trajectory, acceleration vector is not the same with the velocity vector but depends on the change in velocity. The acceleration path is perpendicular to the velocity vector path.

At point 3, since the ball path is parabolic, the velocity at that point is not equal to zero but will tend to take a horizontal path.

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Answer:

answer

Explanation:

The initial horizontal velocity canbe determined by measuring the diameter of the ball and dividing by the time that it takes for the ball to move across the photogate. That is Vo = d/t.

When Vertical velocity refers to the speed of an object's movement in a vertical direction.

Answer:

Work = Force × Distance

Therefore Force = Work/Distance

= 45 joules/3 metres

= 15 newton

Explanation:

Answer:

15

Explanation:

15

The sag in the tensioned cable depends on the weight of the cable and

the applied tension.

Reasons:

The given parameters are;

Height of the towers are; h₁ = 40 m and h₂ = 30 m

Horizontal distance between the towers, l = 300 m

The tension in the conductor, T = 1,590 kg

Weight of the conductor, w = 0.8 kg/m

Required:

The clearance space of the conductor mid-way between the support

Solution:

The distance, x₁, to the lowest point on the conductor, from the tower, h₁, is given by the formula;

[tex]\displaystyle x_1 = \mathbf{ \frac{l}{2} -\frac{T \cdot h}{w \cdot l}}[/tex]

Where;

h = h₂ - h₁ = 40 m - 30 m = 10 m

Which gives;

[tex]\displaystyle x_1 = \frac{300}{2} -\frac{1590 \times 10}{0.8 \times 300} = \mathbf{83.75}[/tex]

The sag at the lowest point is give by the formula;

[tex]\displaystyle d_1 = \mathbf{ \frac{W \cdot x_1^2}{2 \cdot T}}[/tex]

Therefore;

[tex]\displaystyle d_1 = \frac{0.8 \times 83.75^2}{2 \times 1590} = \frac{4489}{2544} \approx 1.765[/tex]

The lowest point sag, s₁ = 30 m - 1.765 m = 28.235 m

The distance of the midpoint from the lowest point, x, is therefore;

[tex]\displaystyle x = \frac{l}{2} - x_1 = \frac{300}{2} - 83.75 = \mathbf{ 66.25}[/tex]

Which gives;

[tex]\displaystyle d_x = \frac{0.8 \times 66.25^2}{2 \times 1590} = \frac{53}{48} \approx 1.104[/tex]

The clearance midway between the support ≈ 28.235 + 1.104 ≈ 29.34

The clearance midway between the support ≈ 29.34 m

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The electromagnetic wave used  in nuclear research and mineral exploration are gamma rays.

Electromagnetic waves refer to those rays that are found in the electromagnetic spectrum. They are collectively called light. Electromagnetic waves are used for different purposes.

Gamma rays are useful in nuclear research especially in identifying decaying radionuclides. Also, gamma rays are used for geologic mapping, mineral exploration, and identification of environmental contamination.

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Answer:

16 to 25 degrees f

Explanation:

Answer:

Explanation:

true

Answer:

seismic waves

Explanation:

i think thats the answer

Answer:

5m/s ez

Explanation:

plz make me brainliest

Answer:  34cm each year

Explanation: