Find the temperature

The grasshopper is 1.47 m far away from the point where it jumps to the point where its lands.

To calculate the distance of the landing point of the grasshopper to the point where its jumps, we use the formula of range.

Range can be defined as the horizontal distance between the point of projection to the point where the projectile hit the plain again.

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, The grasshopper is 1.47 m far away from the point where it jumps to the point where its lands.

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Hi there!

Part 1:

According to Newton's Third Law, every action has an EQUAL and OPPOSITE reaction.

For a collision, the objects involved exert an EQUAL and OPPOSITE impulse on each other, which means the forces exerted are equal as well.

Therefore, the school bus and Toyota Prius exert an EQUAL FORCE on each other.

Part 2:
The school bus has a greater mass.

Part 3:

Using Newton's Second Law:
[tex]\Sigma F = ma[/tex]

Using this relationship, the smaller the mass, the larger the acceleration. Since the forces are EQUAL, the Toyota Prius will experience a larger acceleration because it has a smaller mass.

Answer:

A dot B = C     is the vector equation for this expression

A · B = A B cos θ

3 * 4 cos θ = 1       the value 1 is their dot product

cos θ = 1 / 12 = .083       θ = 85.2 deg

Answer:

encourage all attempts at reading, writing, and speaking

Explanation:

Answer:

Find the change in momentum of the upper stage, that is:

∆p = m(vf - vi)

m being the mass of the upper stage

vf being the final velocity which was given

vi being the initial which was also given

find ∆p

then use ∆p in the same equation

∆p being the answer you got above

m being the mass of the lower stage (given)

vi being the initial velocity (given)

vf being the final velocity of the lower stage which you were asked to find

Explanation:

During a collision the change in momentum (∆p) for both objects is equal regardless of their speeds or masses before or after the equation

[tex]\text{Given that,}\\ \\\text{Power,} ~P=24~ KW = 24000~ W\\\\\text{Force,} ~F=600~N\\\\\text{We know that,}\\\\P=Fv\\\\\implies v = \dfrac{P}{F} = \dfrac{24000}{600}=40~ ms^{-1}\\\\\\\text{It travels at 40 m/s}[/tex]

Answer: what I know that the unit is farad

Explanation:

Try this solution, all the details are in the attachment. note, the answer is marked with orange colour. If it is possible, check the provided solution in other sources.

Answer: ≈0.81c.

Based in the relationship between static frictional force and coefficient of static friction, the force of static friction is 24 N.

Friction is a force that opposes the relative motion of an object moving over another at their surafce of contact.

Frictional force is constant for each type of material. This constant is known as coefficient of friction.

The coefficient of friction is given as follows:

From the data given:

Force of static friction = 0.60 × 40

Force of static friction = 24 N.

Therefore, the force of static friction is 24 N.

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Answer:

thats is so hard

Explanation:

i cant answer that

Answer:

The resistance of a 60 watt light bulb designed to operate at 120 volts is 240 Ω.

Explanation:

Answer:

0.5 mega grams

Explanation:

Answer:

The static frictional force is the force between the magnetic forces statically and contains only two objects during this physical reaction. It results in electricity and a slight shock.

Answer:

D. remain at rest

Explanation:

The magnitude of the kinetic friction force between the crate and the floor is 33.15 N.

The magnitude of kinetic frictional force is determined by applying Newton's second law of motion as shown below;

F - Ff = ma

where;

Subsitute the given parameters and solve for kinectic frictional force

197 - Ff = 49.1(3.337)

197 - Ff = 163.85

Ff = 33.15 N

Thus, the magnitude of the kinetic friction force between the crate and the floor is 33.15 N.

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What the equation given?

Maximum velocity occurs at the equilibrium position

So

Now

Now

As we know the formula

[tex]\\ \rm\rightarrowtail V_max=A\omega[/tex]

These expressions can be used

The maximum speed of the oscillating object will be given by [tex]V_{max}=Aw[/tex]

An oscillation is defined as the repitative periodic motion of any object about its mean or equilibrium position.

The given equation is as follows:

x(t)=Acost

Maximum velocity occurs at the equilibrium position x=0

x(0)=Acos0

x(0)=A

Hence the maximum velocity will be  [tex]V_{max}=Aw[/tex]

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Hi there!

We can begin by summing the forces acting on the car.

Along the axis of the incline, the forces of friction and gravity are working. The force of friction points towards the top of the ramp, while the force of gravity works towards the bottom.

We can use trigonometry to determine the force due to gravity along the ramp.

[tex]F_g = Mg sin\theta[/tex]

The force due to friction is equal to:
[tex]F_f = \mu N[/tex]

The normal force is the vertical component of the force due to gravity, so:
[tex]F_f = \mu Mgcos\theta[/tex]

Now, the combination of these two forces produces a component of the centripetal force. Drawing a diagram, we see that the true centripetal force is the HYPOTENUSE, while these forces sum up to its horizontal component along the ramp.

Therefore:
[tex]F_c sin\theta = Mgsin\theta - \mu Mgcos\theta[/tex]

The centripetal force is equivalent to:
[tex]F_c = \frac{Mv^2}{r}[/tex]

m = mass (kg)

v = velocity (m/s)
r = radius (m)

Rewrite:
[tex]\frac{Mv^2}{r}sin\theta = Mgsin\theta - \mu Mgcos\theta[/tex]

Cancel out 'M'.

[tex]\frac{v^2}{r}sin\theta = gsin\theta - \mu gcos\theta[/tex]

Rearrange to solve for 'r'.

[tex]r = \frac{v^2sin\theta}{gsin\theta - \mu gcos\theta}[/tex]

Plug in values and solve.
[tex]r = \frac{(27.1^2)sin(29.1)}{(9.8)sin(29.1) - 0.4(9.8)cos(29.1)} = \boxed{266.365 m}[/tex]

Answer:

0.98 [sec].

Explanation:

1) the time when the player is moving up equals the time the player falls, it can be 't'. Then

2) the equation of moving down can be written as

[tex]\frac{g*t^2}{2} =h, \ where[/tex]

g=10 [m/s²], t - time, h - the given height.

3) according to the formula above the time spent to moving down is:

[tex]t=\sqrt{\frac{2h}{g}};[/tex]

[tex]t=\sqrt{\frac{2.4}{10}}=\sqrt{0.24} =0.49[s].[/tex]

4) finally, the total time is t*2=0.98[sec].

1st image - B) Object position

2nd image - A) Focal length

3rd image - C) Image position

The athlete can improve performance by building strength, coordination and balance.

This is an individual who is proficient in sports and other forms of physical exercise.

Improvement of performance based on the kinematic variable of speed and acceleration can be achieved by building strength, coordination and balance by performing plyometric exercises etc.

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Biomolecules are fundamental for life. They required both energy supply and building structures.

Biomolecules are organic molecules that fundamental for life and must be ingested from the regular diet.

Biomolecules include proteins (e.g., meal proteins), lipids (fats) and also carbohydrates (e.g., glucose).

Biomolecules required both energy supply (e.g., glucose) and building structures (e.g., amino acids in proteins).

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Answer:

A. An unbalanced force.

Explanation:

There needs to be a net force in order for the nail to be driven into presumably the wall. Without the net force then the hammer and nail wouldn't move.

Answer:

The answer should be choice B.

With the use of first and third equation of motion, the distance covered by a T-Rex is 30.51 m

When a body is in linear motion, the body is moving in a straight line. some of the parameters to consider are:

Given that a T-Rex move from 0 m/s to 9 m/s in 6.78 seconds, the distance covered can be found by calculating the acceleration.

Let us use equation 1

V = U + at

9 = 0 + 6.78a

a = 9 / 6.78

a = 1.33 m/[tex]s^{2}[/tex]

Now let us use equation 3

[tex]v^{2}[/tex] = [tex]u^{2}[/tex] + 2as

[tex]9^{2}[/tex] = 2 x 1.33 x S

81 = 2.655S

S = 81/2.655

S = 30.51 m

Therefore, the distance covered by a T-Rex is 30.51 m.

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Answer:

1716.75 J

Explanation:

Step 1: First check what we are provided with. As per given question we have:

mass (m) = 70 kg, height (h) = 2.5 m and acceleration due to gravity (g) = 9.81 m/s².

Step 2: Check what we are asked to find out.

Work done = Change in Potential energy

The stuff required to solve this question is potential energy. Using the formula: P = mgh. Where P is Potential energy, m is mass, g is acceleration due to gravity and h is height.

Step 3: Substitute the known values in the above formula.

→ P = 70 × 2.5 × 9.81

→ P = 1716.75 J

Hence, the work done against the force of gravity is 1716.75 J.

The magnitude of the average force exerted on the ball by the club is 429.27 N.

Force can be defined as the product of mass and acceleration

To calculate the force exerted on the ball by the club, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, The magnitude of the average force exerted on the ball by the club is 429.27 N.

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Answer:

m = 4.5 kg

Explanation:

w = - 1800 j

Fd = - 1800 j

mgd = - 1800 j

m = - 1800 ÷(gd)

m = - 1800 ÷( 10×-40)

m = 4.5 kg

It is amount of force or motion that cause the movement or moving of body.

Thus it is well explained.

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1.No they do not violate the conservation of law. Because the initial momentum can be zero.

2. The momentum is conserved in both elastic and inelastic collision.

The momentum of the body before the contact is always equal to the momentum of the body after the impact, according to the rule of conservation of momentum.

According to the law of conservation of momentum;

Momentum before collision =Momentum after collision

[tex]\rm m_1u_1 + m_2u_2 = v(m_1 + m_2)[/tex]

1.No they do not violate the conservation of law. Because the initial momentum can be zero.

2. The momentum is conserved in both elastic and inelastic collision. Hence the conservation law works with elastic and inelastic collisions.

3. An elastic collision is one in which kinetic energy is conserved, meaning that the total kinetic energy before and after the impact is the same.

The object bounces once after the body gets struck. When the body's final velocity is zero it follows the stationary object after the collision.

4.. Airbags are meant to inflate automatically in the case of an abrupt deceleration or impact force indicative of a collision.

When the bag is fully inflated, it protects the driver and/or passenger by Increasing cushioning around the driver's neck, head, and spine.

5. Stretchable seatbelts are meant to allow a passenger's considerable velocity to be slowly reduced, preventing the passenger from being flung forward abruptly.

As a result, if the automobile collides, serious casualties can be avoided.

Hence no they do not violate the conservation of law. Because the initial momentum can be zero.

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Explanation:

conduction- putting a metal spoon in fire

convection- boiling water. In ventilation

Radiation- suns heat reaches us by radiation. heat from fire place reaches us by radiation.