who won the battle of Buxar​

Answer:

Explanation:

Here's what we know and in which dimension:

y dimension:

[tex]v_0=30[/tex] m/s

v = 0 (I'll get to that injust a second)

a = -9.8 m/s/s

The final velocity of 0 is important because that's the velocity of the ball right at the very top of its travels. If we knew how long it takes to get to that max height, we can also use that to find out how long it will take to hit the ground. Therefore, we will find the time it takes to reach its max height and pick up with the investigation of what this means after.

x dimension:

Δx = 70 m

v = ??

Velocity is our unknown.

Solving for the time in the y dimension:

[tex]v=v_0+at[/tex] and filling in:

0 = 30 + (-9.8)t and

-30 = -9.8t so

t = 3.1 seconds

We know it takes 3.1 seconds to get to its max height. In order to determine how long it will take to hit the ground, just double the time. Therefore, it will take 6.2 seconds for the ball to come back to the ground, which is where the persom trying to catch the ball comes in. We will use that time in our x dimension now.

In the x dimension, the equation we need is just a glorified d = rt equation since the acceleration in this dimension is 0.

Δx = vt and

70 = v(6.2) so

v = 11.3 m/s

This question is incomplete, the missing diagram is uploaded along this answer below.

Answer:

- the magnitude of the tension force exerted by the hamstring muscles on the leg is 990 N

- the magnitude of compression force at the knee joint is 900 N

Explanation:

Given the data in the question and diagram below;

Net torque = 0

Torque = force × lever arm

so

F[tex]_{ConF[/tex]  × ( 15.0 in + 1.5 in ) = T[tex]_{HonL[/tex] × 1.5 in

given that F[tex]_{ConF[/tex] = 90 N

90 × ( 15.0 in + 1.5 in ) = T[tex]_{HonL[/tex] × 1.5 in

90 N × 16.5 in =  T[tex]_{HonL[/tex] × 1.5 in

T[tex]_{HonL[/tex] = ( 90 N × 16.5 in ) / 1.5 in

T[tex]_{HonL[/tex] = 990 N

Therefore, the magnitude of the tension force exerted by the hamstring muscles on the leg is 990 N

b) magnitude of compression force at the knee joint;

In equilibrium, net force = 0

along horizontal

F[tex]_{FonB[/tex] - T[tex]_{HonL[/tex] + F[tex]_{ConF[/tex] = 0

we substitute

F[tex]_{FonB[/tex] - 990 + 90 = 0

F[tex]_{FonB[/tex] - 900 = 0

F[tex]_{FonB[/tex] = 900 N

Therefore, the magnitude of compression force at the knee joint is 900 N

Answer:

the horizontal component of the force is 50 N

Explanation:

Given;

force applied by the man, F = 100 N

angle of inclination of the force, θ = 60⁰

mass of the dog, m = 20 kg

The horizontal component of the force is calculated as;

[tex]F_x = F\times cos(\theta)\\\\F_x = 100 \ N \times cos(60^0)\\\\F_x = 100\ N \times 0.5\\\\F_x = 50 \ N[/tex]

Therefore, the horizontal component of the force is 50 N

Answer:

The speed of the combined mass after the collision is 2.1 m/s.

Explanation:

mass of runner, m = 70 kg

speed  of runner, u = 2.7 m/s

mass of shortstop, m' = 85 kg

speed  of shortstop, u' = 1.6 m/s

Let the velocity of combined system is v.

Use conservation of momentum

Momentum before collision = momentum after collision

m u + m' u' = (m + m') v

70 x 2.7 + 85 x 1.6 = (70 + 85) v

189 + 136 = 155 v

v = 2.1 m/s

Answer:

559.5 N

Explanation:

Applying,

v² = u²+2gs............. Equation 1

Where v = final velocity,

From the question,

Given: s = 5.10 m, u = 0 m/s ( from rest)

Constant: 9.8 m/s²

Therefore,

v² = 0²+2×9.8×5.1

v² = 99.96

v = √(99.96)

v = 9.99 m/s

As the diver eneters the water,

u = 9.99 m/s, v = 0 m/s

Given: t = 1.34 s

Apply

a = (v-u)/t

a = 9.99/1.34

a = -7.46 m/s²

F = ma.............. Equation 2

Where F = force, m = mass

Given: m = 75 kg, a = -7.46 m/s²,

F = 75(-7.46)

F = -559.5 N

Hence the average force exerted on the diver is 559.5 N

Answer:

12+2=24+30+2=66

Explanation:

Answer:

(a) 9.28 m/s2

(b) 9.03 m/s2

(c) 9.8 m/s2

(d) 450 N, 670 N

Explanation:

mass of elevator + student, m = 870 kg

Reading of scale, R = 450 N

(a) When the elevator goes down, the weight decreases.

Let the acceleration is a.

By the Newton's second law

m g - R = m a

870 x 9.8 - 450 = 870 a

a = 9.28 m/s2

(b) R = 670 N

Let the acceleration is a.

870 x 9.8 - 670 = 870 a

a = 9.03 m/s2

(c) If the scale reads zero, it mean the elevator is falling freely. The acceleration is downwards and its value is 9.8 m/s2.

(d) Tension in cable is 450 N and 670 N.

Answer:

By measure the effective length and the time period of the pendulum.

Explanation:

Let the student take the oscillating pendulum at the planet.

He measure the time period of the pendulum  by using the stop watch or the ordinary watch.

Then measure the effective length of the pendulum which is the distance between the center of gravity of the bob and the point of suspension of the pendulum.

Now, use the formula of the time period of the pendulum,

[tex]T =2\pi\sqrt\frac{L}{g}[/tex]

Here, L is the effective length of the pendulum, g is the acceleration due to gravity at the planet and T is time period of the pendulum.  

By rearranging the terms, we get

[tex]T =2\pi\sqrt\frac{L}{g}\\\\T^{2}=4\pi^2\times\frac{L}{g}\\\\g =\frac{4\pi^2L}{T^2}[/tex]

Here, by substituting the values of L and T, the student get the value of acceleration due to gravity at that planet.

Answer:

The efficiency of Carnot's heat engine is 26.8 %.

Explanation:

Temperature of hot reservoir, TH = 100 degree C = 373 K

temperature of cold reservoir, Tc = 0 degree C = 273 K

The efficiency of Carnot's heat engine is

[tex]\eta = 1-\frac{Tc}{T_H}\\\\\eta = 1 -\frac{273}{373}\\\\\eta = 0.268 =26.8 %[/tex]

The efficiency of Carnot's heat engine is 26.8 %.

The boy pushed the sled for 16 seconds.

We have a boy who pushes his little brother on a sled.

We have to determine for how long time does boy push the sled.

The Work - Energy theorem states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.

According to the question -

The sled is initially at rest → initial velocity (u) = 0.

Final velocity (v) = 4 m/s

Mass of boy and sled (M) = 40 kg

Power developed (P) = 20 W = 20 Joules/sec

According to work - energy theorem -

Work done (W) = Δ E(K) = E(f) - E(i)

Therefore -

W = ([tex]\frac{1}{2} \times 40 \times 4 \times 4 - \frac{1}{2}[/tex] x 40 x 0) = 320 Joule

Now, Power is defined as the rate of doing work -

P = [tex]\frac{dW}{dt}[/tex] = [tex]\frac{W}{t}[/tex]

20 = [tex]\frac{320}{t}[/tex]

t = 16 seconds

Hence, the boy pushed the sled for 16 seconds.

To solve more questions on Work, Energy and Power, visit the link below -

https://brainly.com/question/208670

#SPJ2

Answer:

M L1 = m L2       torques must be zero around the fulcrum

M = m L2 / L1 = .3 kg * 28 cm / 22 cm = .382 kg

Answer:

So the answer is B. A is wrong because negative answer = deceleration

Answer:

The correct answer is "4000 J".

Explanation:

Given that,

Force,

= 200 N

Displacement,

= 20 m

Now,

The work done will be:

⇒ [tex]Work=Force\times displacement[/tex]

By putting the values, we get

              [tex]=200\times 20[/tex]

              [tex]=4000 \ J[/tex]

Answer:

Energy Bill fluctuations are inevitable and depend on a variety of different factors. Two of the most important are the current weather your home is experiencing and the current price per Kilowatt Hour (which fluctuates more than you might think).

Answer: The weight of the object is 29.4 N

Explanation:

To calculate the weight of the object, we use the equation:

[tex]W=m\times g[/tex]

where,

m = mass of the object = 3 kg

g = acceleration due to gravity = [tex]9.8m/s^2[/tex]

Putting values in above equation, we get:

[tex]W=3kg\times 9.8m/s^2\\\\W=29.4N[/tex]

Hence, the weight of the object is 29.4 N

Answer:

A

Explanation:

I guess not that much confidential!

We call the acceleration of an object moving in uniform circular motion— resulting from a net external force—the centripetal ...

Answer:

0.444atm

Explanation:

Using the combined gas law equation;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (

P2 = final pressure (

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to this question,

P1 = 101.3 kPa = 101.3 × 0.00987 = 0.999atm

P2 = ?

V1 = 80L

V2 = 160L (double of V1)

T1 = 34°C = 34 + 273 = 307K

T2 = 0°C = 0 + 273 = 273K

Using P1V1/T1 = P2V2/T2

0.999 × 80/307 = P2 × 160/273

79.92/307 = 160P2/273

Cross multiply

307 × 160P2 = 79.92 × 273

49120P2 = 21818.16

P2 = 21818.16 ÷ 49120

P2 = 0.444

P2 = 0.444atm

Answer:

C. both a speed and a direction

Answer:

Refer to the attachment!~

Answer:

the electric field is  3.91 x 10⁶ N/C

Explanation:

Given the data in the question;

Electric field at a point due to point charge is;

E = kq/r²

where k is the constant, r is the distance from centre of terminal to point where electric field is, q is the excess charge placed on the centre of terminal of Van de Graff,a generator

Now, given that r = 3.9 m, k = 9.0 x 10⁹ Nm²/C², q = 6.60 mC = 6.60 x 10⁻³ C

so we substitute into the formula

E = [(9.0 x 10⁹ Nm²/C²)( 6.60 x 10⁻³ C)] / ( 3.9 )²

E = 59400000 / 15.21

E = 3.91 x 10⁶ N/C

Therefore, the electric field is  3.91 x 10⁶ N/C

Answer:

  T = 9056 K

Explanation:

In the exercise they indicate that the body can be approximated by a black body, for which we can use the Wien displacement relation

                 λ T = 2,898 10⁻³

where lam is the wavelength of the maximum emission

                T = 2,898 10⁻³ /λ

let's calculate

                 T = 2,898 10⁻³ / 320 10⁻⁹

                  T = 9.056 10³ K

                  T = 9056 K

Answer:

Force = 1.632 Newton

Explanation:

Given the following data;

Pressure = 102 kPa

Area = 0.016 m²

To find what force the atmosphere exert on the palm of your hand;

Mathematically, pressure is given by the formula;

[tex] Pressure = \frac {Force}{area} [/tex]

Force = 102 * 0.016

Force = 1.632 Newton

Answer:

Mm, thats the answer trust me men

Explanation:

Energy input = F×d = (150 N)(5.5 m) = 825 J

Energy output = mgh = (66 kg)(9.8 m/s^2)(1.10 m) = 711 J

efficiency = [tex]\dfrac{\text{output}}{\text{input}}[/tex]×100% = 86.2%

Answer:

The right solution is "50200 days".

Explanation:

Given:

Calories intake,

= 6000 kcal,

or,

= [tex]2.52\times 10^7 \ J[/tex]

Force,

= 500 N

As we know,

⇒ [tex]Work \ done = Force\times distance[/tex]

Or,

⇒ [tex]distance = \frac{Work \ done}{Force}[/tex]

By putting the values, we get

                  [tex]=\frac{2.52\times 10^7}{500}[/tex]

                  [tex]=0.502\times 10^5[/tex]

                  [tex]=50200 \ m[/tex]

hence,

The number of days will be:

= [tex]\frac{50200}{1}[/tex]

= [tex]50200 \ days[/tex]