A car's speed changes from 10m/s to 35m/s in 15.0s along a straight road while moving in one way. Find the magnitude of the acceleration of the car during this period.​

Answer:

the magnetic force on the proton is zero.

Explanation:

Given;

speed of the proton, v = 10 m/s

magnitude of the magnetic field, B = 3 T

The magnitude of the magnetic force on the particle is calculated as;

F = qvBsinθ

where;

θ is the angle between the velocity of the particle and the magnetic field

Since the particle is moving parallel to the magnetic field, θ = 0

F = qvBsin(0)

F = 0

Therefore, the magnetic force on the proton is zero.

Answer:

This is a trick question, in that the numbers do not matter.

Study the dependence of the magnetic force on the direction of the magnetic field and the direction of motion of the particle. When is it a maximum? When is it a minimum?

Hint: In vector notation, this is often expressed as q v x B, where q is the electric charge of the particle, v is its velocity (a vector), and B is the magnetic field vector.

Answer:

98Kg

Explanation:

Explanation:

Hi

Hi

Hi

Hi

BRAINILIEST PLEASE

What is 4+4+2+2

.The answer is 12

Hope this helps

Answer:

[tex]d=1200\ lb/ft^3[/tex]

Explanation:

Given that,

The weight of the statue, m = 96 lb

The volume of the statue = 0.08 ft³

We need to find the statue’s weight density. We know that the density of an object is the mass of an object divided by its volume. So,

[tex]d=\dfrac{m}{V}\\\\d=\dfrac{96\ lb}{0.08\ ft^3}\\\\d=1200\ lb/ft^3[/tex]

So, the density of the statue is equal to [tex]1200\ lb/ft^3[/tex].

Answer:

Explanation:

the block will move to the right side with small velocity because the force from the left side greater than force from right side. Velocity will be less because of friction and gravitational attraction.

If a block of mass 2 kg is acted upon by two forces: 3 N (directed to the left) and 4 N (directed to the right), then the block would move towards the right side.

Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change of momentum.

As given in the problem statement If a block of mass 2 kg is acted upon by two forces: 3 N (directed to the left) and 4 N (directed to the right),

The net force acting on the block = 4 Newtons - 3 newtons

                                                       = 1 Newton

Thus, we can say that the block would move to the right side.

To learn more about Newton's second law here, refer to the link given below;

brainly.com/question/13447525

#SPJ2

Answer:

Hydrofluorocarbons (HFCs) are man-made organic compounds that contain fluorine and hydrogen atoms, and are the most common type of organofluorine compounds. Most are gases at room temperature and pressure.

Answer:

Hydrofluorocarbons (HFC) are man-made organic compounds that contain fluorine and hydrogen atoms and are the most common type of organofluorine compounds. Most are gases at room temperature and pressure. HFCs are produced synthetically and are used primarily as refrigerants. In general, they are relatively non-flammable, chemically stable, and nonreactive.

Answer:

i think its oil

Explanation:

It was made & used as early as the fourth century BC.

Answer:

the gain in gravitational potential energy is 0.4369 J

Explanation:

Given the data in the question;

from the definition of density, we know that;

ρ = m / V

where ρ is density, m is the mass and V is the volume.

Now, lets make mass the subject of the formula

m = ρV  ------ let this be equation 1

Now, we know that potential energy PE = mgh ------- let this be equation 2

where m is mass, g is acceleration due gravity, and h is the height.

substitute equation 1 into 2

PE = ρVgh

given that; V =  110 mL = 110 × 10⁻⁶ m³, h = 38.6 cm = 38.6 × 10⁻² m, g = 9.8 m/s², ρ = 1.05 × 10³ kg/m³

we substitute

PE = (1.05 × 10³ kg/m³) × (110 × 10⁻⁶ m³) × 9.8 m/s² × 38.6 × 10⁻² m

PE =  0.4369 J

Therefore,  the gain in gravitational potential energy is 0.4369 J

Answer:

d = 4 d₀o

Explanation:

We can solve this exercise using the relationship between work and the variation of kinetic energy

         W = ΔK

In that case as the car stops v_f = 0

the work is

          W = -fr d

we substitute

          - fr d₀ = 0 - ½ m v₀²

           d₀ = ½ m v₀² / fr

now they indicate that the vehicle is coming at twice the speed

          v = 2 v₀

using the same expressions we find

           d = ½ m (2v₀)² / fr

           d = 4 (½ m v₀² / fr)

           d = 4 d₀o

Answer:

[tex]W=2.69*10^{6}J[/tex]

Explanation:

Generally the equation for momentum is mathematically given by

Diameter [tex]d=10m[/tex]

Radius [tex]r =\frac{d}{2}=>5m[/tex]

Height [tex]h=4m[/tex]

Depth [tex]d=3m[/tex]

Generally the equation for Area of the circle is mathematically given by

[tex]A=\pi r^2[/tex]

[tex]A=\pi 5^2[/tex]

[tex]A=78.5m^2[/tex]

Generally the equation for Work done is mathematically given by

[tex]d_w=d_f*x[/tex]

Where

[tex]d_f=(\rho*d_v)*g[/tex]

Therefore

[tex]d_w=((\rho*d_v)*g)*x[/tex]

[tex]d_w=((1000*78.5)*9.8)*x dx[/tex]

[tex]d_w=769300x dx[/tex]

[tex]W=\int dw[/tex]

[tex]\int dw=769300 \int^4_{3}x dx[/tex]

[tex]W=(769300\2)*7[/tex]

[tex]W=2.69*10^{6}J[/tex]

Answer:

714.286g

Explanation:

Q=mcT

60000=m×4.2×(40-20)

m=714.285....g

Answer:

[tex]T=728.9K[/tex]

Explanation:

Power [tex]P=100W[/tex]

Diameter [tex]d=5[/tex]

Radius [tex]r=2.5cm=>2.5*10^{-2}m[/tex]

Emissivity [tex]e=0.8[/tex]

Generally the equation for Area of Spherical bulb is mathematically given by

[tex]A=4\pi r^2[/tex]

[tex]A=4\pi (2.5*10^{-2}m)^2[/tex]

[tex]A=7.85*10^{-3}m^2[/tex]

Generally the equation for Emissive Power bulb is mathematically given by

[tex]E=e\mu AT^4[/tex]

Where

[tex]\mu=Boltzmann constants\\\\\mu=5.67*10^{-8}[/tex]

Therefore

[tex]T^4=\frac{E}{e\mu A}[/tex]

[tex]T^4=\frac{100}{0.8*5.67*10^{-8}*7.85*10^{-3}m^2}[/tex]

[tex]T=^4\sqrt{2.80*10^{11}}[/tex]

[tex]T=728.9K[/tex]

256.2 is the answer.

Here, we are asked to calulate the electric field.

Using the following formula:

E=k*lql/r^2

Yeah! Something that can easily bamboozle you.

So:

E=electric field

k=constant (8.99x10^9)

q=charge

ll= absolute value marks ( I still remember from my Algebra Course))

r=distance

Now, let's plug in the numbers.

E=8.99x10^9*l-5.77x10^-9l/(0.450)^2

And  yes! That's a LOT!

But I solved the same problem on paper 2 minutes ago:)

If you perform the calculations, you will get 256.2

Well. THIS one was correct or Acellus.

So our answer is:

Answer:

15.44 m/s

Explanation:

Below is the given values:

The bell in rings with a sound = 474 Hz

If the frequency of pigeon hears = 453 Hz

Speed = ?

Use below formula:

Frequency = [(Vs - Vo) / Vs ] x fo

Vs = Speed of sound

Vo = Speed of observer

fo = Sound frequency

Frequency = [(Vs - Vo) / Vs ] x fo

453 = [(343 - Vo) / 343 ] x 474

453 / 474 =  [(343 - Vo) / 343 ]  

0.955 = (343 - Vo) / 343

0.955 x 343 = 343 - Vo

327.56 = 343 - Vo

Vo = 343 - 327.56

Vo = 15.44 m/s

This question is incomplete, the complete question is;

Consider the cylindrical weir of diameter 3m and length 6m. If the fluid on the left has a specific gravity of 1.6 and on the right has a specific gravity of 0.8, Find the magnitude and direction of the resultant force.

Answer:

- the magnitude of the resultant force is 557.32 kN

- the direction of resultant force is  48.29°

Explanation:

Given the data in the question and the diagram below,

First we work on the force on the left hand side.

Left Horizontal

[tex]F_{LH[/tex] = βgAr

here, h = 3/2 = 1.5 m, β = 1.6, g = 9.81 m/s², A = 3 m × 6 m = 18 m²

we substitute

[tex]F_{LH[/tex] = βgAh = ( 1.6 × 1000 ) × 9.81 × 18 × 1.5 = 423792 N

Left Vertical

[tex]F_{LV[/tex] = ( βgπh² / 2 ) × W

we substitute

[tex]F_{LV[/tex] = [ ( ( 1.6 × 1000 ) × 9.81  × π(1.5)² ) / 2 ] × 6 = 332845.458 N

Now we go to the right hand side

Right Horizontal

[tex]F_{RH[/tex] = βgAh

here, h' = 1.5/2 = 0.75 m, β = 0.8, g = 9.81 m/s², A = 1.5 m × 6 m = 9 m²

we substitute

[tex]F_{RH[/tex] = ( 0.8 × 1000 ) × 9.81 × 9 × 0.75 ) = 52974 N

Right Vertical

[tex]F_{RV[/tex] = ( βgπh² / 4 ) × W

we substitute

[tex]F_{RV[/tex] = [ ( ( 0.8 × 1000 ) × 9.81  × π(1.5)² ) / 4 ] × 6 =  83211.36 N

Hence

Fx = [tex]F_{LH[/tex] - [tex]F_{RH[/tex] = 52974 N - 423792 N =  370818 N

Fy = [tex]F_{LV[/tex] + [tex]F_{RV[/tex] = 332845.458 N + 83211.36 N = 416056.818 N

R = √( Fx² + Fy² ) = √[ (370818 N)² + (416056.818 N)² ] = 557323.3 N

R = 557.32 kN

Therefore, the magnitude of the resultant force is 557.32 kN

Direction of resultant force;

tanθ = Fy / Fx

we substitute

tanθ = 416056.818 N / 370818 N

tanθ = 1.121997

θ = tan⁻¹( 1.121997 )

θ = 48.29°

Therefore, the direction of resultant force is  48.29°

Answer:

a. 13.96 m/s in the +x direction

b. -1.76 m/s that is 1.76 m/s in the -x direction

Explanation:

(a) Calculate the magnitude of the velocity of the puck after a force of 37.0 N directed to the right has been applied for 0.050 s.

Since Impulse, I = Ft = m(v₂ - v₁) where F = force applied = + 37.0 N(positive since it is applied to the right), t = duration of applied force = 0.050 s, m = mass of hockey puck = 0.160 kg, v₁ = initial velocity of hockey puck =  + 2,40 m/s (positive since it is moving to the right) and v₂ = final velocity of hockey puck

So, making v₂ subject of the formula, we have

v₂ = v₁ + Ft/m

substituting the values of the variables into the equation, we have

v₂ = v₁ + Ft/m

v₂ = 2.40 m/s + 37.0 N × 0.050 s/0.160 kg

v₂ = 2.40 m/s + 1.85 Ns/0.160 kg

v₂ = 2.40 m/s + 11.56 m/s

v₂ = 13.96 m/s

(b) If instead, a force of 13.3 N directed to the left is applied from t = 0 to t =0.050 s, what is the final velocity of the puck?

Since Impulse, I = Ft = m(v₂ - v₁) where F = force applied = - 13.3 N(positive since it is applied to the right), t = duration of applied force = 0.050 s, m = mass of hockey puck = 0.160 kg, v₁ = initial velocity of hockey puck =  + 2,40 m/s (positive since it is moving to the right) and v₂ = final velocity of hockey puck

So, making v₂ subject of the formula, we have

v₂ = v₁ + Ft/m

substituting the values of the variables into the equation, we have

v₂ = v₁ + Ft/m

v₂ = 2.40 m/s + (-13.3 N) × 0.050 s/0.160 kg

v₂ = 2.40 m/s - 0.665 Ns/0.160 kg

v₂ = 2.40 m/s - 4.156 m/s

v₂ = -1.756 m/s

v₂ ≅ -1.76 m/s

(a). The velocity of the puck is 13.96 m/s in the +x direction

(b). The final velocity of the puck is -1.76 m/s which is 1.76 m/s in the -x direction

The velocity of the object is defined as the movement of the object in a particular direction with respect to time.

(a) Calculate the magnitude of the velocity of the puck after a force of 37.0 N directed to the right has been applied for 0.050 s.

It is given that

Impulse, I = Ft = m(v₂ - v₁)

where F = force applied = + 37.0 N(positive since it is applied to the right),

t = duration of applied force = 0.050 s,

m = mass of hockey puck = 0.160 kg,

v₁ = initial velocity of hockey puck =  + 2,40 m/s (positive since it is moving to the right)  

v₂ = final velocity of a hockey puck

So, making v₂ subject of the formula, we have

[tex]V_2=v_1+\dfrac{Ft}{m}[/tex]

substituting the values of the variables into the equation, we have

[tex]v_2=2.40+\dfrac{37\times 0.050}{0.160}[/tex]

[tex]v_2=2.40+11.56[/tex]

[tex]v_2=13.96\ \dfrac{m}{s}[/tex]

(b) If instead, a force of 13.3 N directed to the left is applied from t = 0 to t =0.050 s, what is the final velocity of the puck?

Since Impulse, I = Ft = m(v₂ - v₁)

where F = force applied = - 13.3 N(positive since it is applied to the right),

t = duration of applied force = 0.050 s,

m = mass of hockey puck = 0.160 kg,

v₁ = initial velocity of hockey puck =  + 2,40 m/s (positive since it is moving to the right)  

v₂ = final velocity of a hockey puck

So, making v₂ subject of the formula, we have

[tex]v_2=v_1+\dfrac{Ft}{m}[/tex]

substituting the values of the variables into the equation, we have

[tex]v_2=2.40+\dfrac{-13.3\times 0.050}{0.160}[/tex]

[tex]v_2=2.40-4.156[/tex]

[tex]v+2=-1.76\ \frac{m}{s}[/tex]

Thus

(a). The velocity of the puck is 13.96 m/s in the +x direction

(b). The final velocity of the puck is -1.76 m/s which is 1.76 m/s in the -x direction

To know more about velocity follow

https://brainly.com/question/4931057

Answer: Water fills the bucket until its force/weight is greater than the block’s. The lever tilts over, causing the bucket to water the plant.

Answer: [tex]107.8\ J[/tex]

Explanation:

Given

Initial position of object is (4.4 i+5 j)

Final position of object is (11.6 i -2 j)

Force acting (4i-9j)

Work done is given by

[tex]\Rightarrow W=F\cdot dx\\\Rightarrow W=(4i-9j)\cdot (11.6i-4.4i-2j-5j)\\\Rightarrow W=(4i-9j)\cdot (7.2i-7j)\\\Rightarrow W=28.8+63\\\Rightarrow W=91.8\ J[/tex]

Initial kinetic energy

[tex]K_i=\dfrac{1}{2}\times 2\times 4^2\\\\K_i=16\ J[/tex]

Change in kinetic energy is equal to work done by object

[tex]\Rightarrow K_f=K_i+W\\\Rightarrow K_f=16+91.8\\\Rightarrow K_f=107.8\ J[/tex]

Answer:

27.44 J

Explanation:

We can find the energy at the top of the slide by using the potential energy equation:

At the top of the slide, the swimmer has 0 kinetic energy and maximum potential energy.

The swimmer's mass is given as 7.00 kg.

The acceleration due to gravity is 9.8 m/s².

The (vertical) height of the water slide is 0.40 m.

Substitute these values into the potential energy equation:

Since there is 0 kinetic energy at the top of the slide, the total energy present is the swimmer's potential energy.

Therefore, the answer is 27.44 J of energy when the swimmer is at the top of the slide.

Answer:

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Explanation:

potang ina mooooooo bubu hayop kaaaaapestrng yawa

Explanation:

peste kakkkkaaaaaaaa bubu pesteng yawaaaayaka kaayu kaaaaaaa

Answer:what the choices

Explanation:

Answer:

the tension force of the string on the stone is 30 N

Option d) 30 N is the correct answer.

Explanation:

Given the data in the question;

mass m = 0.2 kg

radius r = 0.6 m

θ = 150 revolutions = 300π rad

time t = 60 seconds

we know that; Angular speed ω = θ / t

we substitute

ω = 300π / 60

ω = 5π rad

Linear speed of stone u = ω × r

we substitute

u = 5π × 0.6

u = 3π m/s

The tension force of the string on the stone is equal to centripetal force, which aid it move in circle;

so

T = mv² / r

we substitute

T = [ 0.2 × (3π)² ] / 0.6

T = 17.7652879 / 0.6

T = 29.6 ≈ 30 N

Therefore, the tension force of the string on the stone is 30 N

Option d) 30 N is the correct answer.

This is the answer hope it helps

Answer: sea-level rise, extreme weather events, and drought, and water scarcity.

Explanation:

Answer:

The correct answer is - it is an error in the observation.

Explanation:

Backlash error is the error in the motion that takes place during shifting the direction of gears. A backlash error is an error in the observation which occurs due to the wear and tear of threads of the screw observed that takes place at the time of reversing the direction of rotation of the thimble where the tip of the screw does not move in the opposite direction but remains stationary for a part of the rotation.

Avoid backlash error:

While taking measurements, the screw should be rotated in one direction only.