plz help will mark the brainliest​

Answer:

If the body is in equilibrium the two forces add up and the third is the opposite of the resultant.

F(1x)=F(1)=10 N

F(2x)=F(2)cos60=5•0.5=2.5 N

F(2y) =F(2)sin60 = 5•0.866= 4.33 N

F(3x) =- F(x)=- (10+2.5 )= -12.5 N

F(3y) =- F(2y)= - 4.33 N

F(3) = sqrt{ F(3x)²+F(3y)²} =13.23 N

tan φ = F(3y)/F(3x) =4.33/12.5=0.364

φ = 19.1⁰ (south-west)

Explanation:

Answer:

A car

Explanation:

Answer:

Please find attached file

Explanation:

The rate of change of velocity with respect to time is known as acceleration. The acceleration of the ball will be 10m/s.

The rate of change of velocity with respect to time is known as acceleration. According to Newton's second law, the eventual effect of all forces applied to a body is its acceleration.

The pace at which a body's velocity varies is represented by acceleration, which is a vector quantity.

The given data in the problem is given by ;

u is the initial speed of ball = 0 m/sec

v is the final speed of the ball= 10 m/sec

t is the time interval= 2 second

a is the acceleration of ball=? m/sec²

The formula for acceleration is;

[tex]\rm a=\frac{(v-u)}{t} \\\\ \rm a=\frac{(20-0)}{2} \\\\\rm a=10\ m/sec^2[/tex]

Hence the acceleration of the ball will be 10m/s².

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Answer:

really fast? xP

Explanation:

Answer: c

Explanation: for #5 it can still be moving, just at a constant speed; therefore, it will be c.

If an object has an acceleration of 0 m/s2, then one can be sure that the object is not changing velocity.

If car A passes car B, then car A must be moving faster than car B, but not necessarily accelerating.

The not consistent with a car that is accelerating is a car is moving at a high speed.

A fullback is running down the football field in a straight line. He starts at the 0-yard line at 0 seconds. At 1 second, he is on the 10-yard line; at 2 seconds, he is on the 20-yard line; at 3 seconds, he is on the 30-yard line; and at 4 seconds, he is on the 40-yard line. This is evidence that he is moving with a constant speed (on average).

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Plot a graph, constant speed of movements represented as straight lines. Time on X axis (seconds)  and position on Y axis (metres) , calculated by moving, e.g 4metres in a second.

Answer:

Explanation:

This is a fast plane! At it's takeoff speed, it could fly around the world in 5 seconds!

Acceleration is defined as the change in speed over the change in time.

[tex]a = \frac{v_f - v_o}{t}[/tex]

Where v_f is the final velocity, v_o is the original velocity, and t is the time.

We know v_f = 8820km/s.

v_o = 0 km/s. The plane "starts at rest before it's engine start." So we know it starts from zero velocity.

t = 32.8 seconds.

You can solve the equation from there.

Answer:

a) The bullet will penetrate 0.375 meters into the bag of sand.

b) The average force exerted by the sand is 15 newtons.

Explanation:

The statement is incorrectly written. The correct form is presented below:

A bullet of mass 50 grams travelling with a speed of 15 meters per second penetrates into a bag of sand and is uniformly brought to rest in 0.05 second.

a) How far the bullet will penetrate into the bag of sand?

b) The average force exerted by the sand?

b) The average force exerted by the sand on the bullet ([tex]F[/tex]), measured in newtons, is determined by Principle of Linear Momentum Conservation and Impulse Theorem:

[tex]F = \frac{m\cdot (v_{f}-v_{o})}{\Delta t}[/tex] (1)

Where:

[tex]m[/tex] - Mass of the bullet, measured in kilograms.

[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speeds of the bullet, measured in meter per second.

[tex]\Delta t[/tex] - Impact time, measured in seconds.

If we know that [tex]m = 0.05\,kg[/tex], [tex]v_{o} = 15\,\frac{m}{s}[/tex], [tex]v_{f} = 0\,\frac{m}{s}[/tex] and [tex]\Delta t = 0.05\,s[/tex], then the average force exerted by the sand is:

[tex]F = \frac{(0.05\,kg)\cdot \left(0\,\frac{m}{s}-15\,\frac{m}{s} \right)}{0.05\,s}[/tex]

[tex]F = -15\,N[/tex]

The average force exerted by the sand is 15 newtons.

a) The distance travelled by the bullet ([tex]\Delta s[/tex]), measured in meters, is determined by application of Principle of Energy Conservation and Work-Energy Theorem:

[tex]\Delta s = \frac{m\cdot (v_{f}^{2}-v_{o}^{2})}{2\cdot F }[/tex] (2)

If we know that [tex]m = 0.05\,kg[/tex], [tex]v_{o} = 15\,\frac{m}{s}[/tex], [tex]v_{f} = 0\,\frac{m}{s}[/tex] and [tex]F = -15\,N[/tex], then the distance travelled by the bullet is:

[tex]\Delta s = \frac{(0.05\,kg)\cdot \left[\left(0\,\frac{m}{s} \right)^{2}-\left(15\,\frac{m}{s} \right)^{2}\right]}{2\cdot (-15\,N)}[/tex]

[tex]\Delta s = 0.375\,m[/tex]

The bullet will penetrate 0.375 meters into the bag of sand.

Answer:

the bonds that are formed

Explanation:

Energy is absorbed in order to break bonds.  Energy is released when new bonds are formed. Bond-making is a process of exothermic. Whether the reaction is endothermic or exothermic refers to the difference between the energy needed to break bonds and the energy released when new bonds are formed.

Answer:

I think its A

Explanation:

Answer:

1. Endurance activities

Only aerobic source

2. Can contract repeatedly

Speed strength and power.

Both aerobic and anaerobic sources

3. Fast contraction speed

Only anaerobic sources

Produce most force

Fatigue quickly

Speed, strength and power

Explanation:

I hope this helps. Although, I'm about 80% sure of the answer.

Answer:

answer is 700 joules

Explanation:

work is equal to force into displacement

W=12.5*56

700 joules

Answer:

The answer should be the one on the top in the picture.

Explanation:

Friction is opposing the downward movement of the crate therefore the friction force is pointing upwards. that is why the answer should be the one on the top of the picture.

Answer:

The average speed of the blood in the capillaries is 0.047 cm/s.

Note: The question is incomplete. The complete question is as follows:

The radius of the aorta is about 1 cm and the blood flowing through it has a speed of about 30 cm/s. Calculate the average speed of the blood in the capillaries given the total cross section of all the capillaries is about 2000 cm².

Explanation:

From the given values:

radius of the aorta, r₁ = 1 cm

speed of blood, v₁ = 30 cm/s

Area of the aorta, A₁ = πr₁² where π = 3.142

Area of aorta = 3.142 × (1)² = 3.142 cm²

Area of the capillaries, A₂ = 2000 cm²

let the average speed of the blood in the capillaries = v₂

From the continuity equation of fluid flow, the product of cross-sectional area of the pipe and the fluid speed at any point along the pipe is always constant. In formula, A₁v₁ = A₂v₂

Using the continuity equation, the average the average speed of the blood in the capillaries can be calculated thus:

A₁v₁ = A₂v₂

v₂ = (A₁v₁) / (A₂)

v₂ = (3.142 x 30) / (2000)

v₂ = 0.047 cm/s

Therefore, the average speed of the blood in the capillaries is 0.047 cm/s.

Answer:

MRI and Radio telescopes

MRI and radio telescopes would be unaffected by satellites that are not working properly.

The longest wavelengths in the electromagnetic spectrum are found in radio waves, which typically have frequencies of 300 gigahertz (GHz) and lower. The wavelength for 300 GHz is 1 mm, which is shorter than a grain of rice, while the wavelength for 30 Hz is 10,000 km (6,200 miles) (longer than the radius of the Earth).

Radio waves move at the speed of light in a vacuum, as do all electromagnetic waves, and at a nearly identical speed in the Earth's atmosphere.

Among the given options, space communication and television broadcasting television broadcasting and GPS  and  radio telescopes and space communication​ are affected by satellites that are not working properly because they are operated by satellite technology.

Hence, only MRI and radio telescopes would be unaffected by satellites that are not working properly.

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Answer:

[tex] \boxed{q = 1.2 \times {10}^{ - 6} C}[/tex]

Explanation:

[tex]f_e = \frac{{q}^{2}k }{ {r}^{2} } \\ q = \sqrt{ \frac{f_e( {r}^{2} )}{k} } = \sqrt{ \frac{5.2(5 \times {10}^{ - 2} )^{2} }{9 \times {10}^{9} } } \\ q =\sqrt{ \frac{5.2(5 \times {10}^{ - 2} )^{2} }{9 \times {10}^{9} } } = \sqrt{ \frac{0.013}{9 \times {10}^{9} } } \\ q = 1.2 \times {10}^{ - 6} [/tex]

Distance covered = (speed) x (time)

Time = (Distance covered) / (speed)

Time = (48 km) / (32 km/hr)

Time = (48/32) hr

Time = 1.5 hrs (1 hour 30 minutes)

Answer:

The frequency of the phonograph record is 0.2 Hz

Explanation:

The frequency of an object moving in uniform circular motion is the number of completed cycles the object makes in a specified time period

The given parameters of the phonograph record are;

The radius of the record = 0.15 m

The number of times the phonograph record rotates, n = 18 times

The time it takes the phonograph record to rotate the 18 times, t = 90 seconds

The frequency of the phonograph record, f = (The number of times the phonograph record rotates) ÷ (The time it takes the phonograph record to rotate the 18 times)

∴ The frequency of the phonograph record, f = n/t = 18/(90 s) = 0.2 Hz

The frequency of the phonograph record = 0.2 Hz.

Answer:

0.3 hour

Explanation:

From the question given above, the following data were obtained:

Distance travelled = 90 km

Speed of 1st car = 50 km/h

Speed of 2nd car = 60 km/h

Time interval separating the two cars =?

Next, we shall determine the time taken for each car to get to the destination. This can be obtained as follow:

For the 1st car:

Distance travelled (d) = 90 km

Speed of 1st car (S₁) = 50 km/h

Time (t₁) =?

Speed = Distance /time

S₁ = d/t₁

50 = 90 / t₁

Cross multiply

50 × t₁ = 90

Divide both side by 50

t₁ = 90 / 50

t₁ = 1.8 hours

For the 2nd car:

Distance travelled (d) = 90 km

Speed of 2nd car (S₂) = 60 km/h

Time (t₂) =?

Speed = Distance /time

S₂ = d/t₂

60 = 90 / t₂

Cross multiply

60 × t₂ = 90

Divide both side by 60

t₂ = 90 / 60

t₂ = 1.5 hours

Finally, we shall determine the time interval separating the arrival of the two cars. This can be obtained as follow:

Time for the 1st car (t₁) = 1.8 hours

Time for the 2nd (t₂) = 1.5 hours

Time interval separating the two cars (tₛ) =?

tₛ = t₁ – t₂

tₛ = 1.8 – 1.5

tₛ = 0.3 h

Thus, the interval separating the arrival of the two cars is 0.3 hour

Answer:

D. Exothermic, because energy is being absorbed from the surroundings

Explanation:

This is true about the Exothemic reaction due to the fact that, the reaction occurs outside the body. During this reaction, the energy being absorbed from the surrounding environment will hit the body surface thereby creating the coldness due to the heat given out from the body  being minimal.

Answer:

Endothermic, because energy is being absorbed from the surroundings.

Explanation:

Answer:

78j

Explanation:

78.125J to be exact

Answer:

C. Sources of electrical current!

Hope this helps :)

Answer:

Sources of electrical current

Explanation:

Got it right on the test.

Answer:

c

Explanation:

Explanation:

Average Speed.=Total distance/Total time=800/10=80m/hour

Answer:

It is not still true that the average velocity of the particle is equal to half the sum of the initial and final velocities when the acceleration of the particle is not constant

Explanation:

The motion of a particle under constant acceleration, 'a', is be given by the following kinematic equations;

v² = u² + 2·a·s

v = u + a·t

Where;

v = The final velocity of the particle

u = The initial velocity of the particle

a = The acceleration of the particle

s = The distance through which the particle travels

t = The time of motion of the particle

By simplifying the above equation, we have;

v² - u² = 2·a·s

(v² - u²)/(2·a) = s

(v - u) × (v + u)/(2 × a) = s

((v - u)/a) × ((v + u)/2) = s

From v = u + a·t, we have;

t = (v - u)/a

∴ ((v - u)/a) × ((v + u)/2) = t × ((v + u)/2) = s

∴ ((v + u)/2) = s/t

The average velocity = (Total distance traveled by the particle) ÷ (The time of travel of the particle)

∴ The average velocity = s/t = ((v + u)/2) = Half the sum of the initial and final velocity

However, it is not still true that the average velocity of the particle is equal to half the sum of the initial and final velocities when the acceleration of the particle, 'a', is not constant, as the velocity time graph is no longer a straight line graph and the distance traveled by the particle, 's', which is the area under the velocity time graph, 'A', (given by the sum of area of the triangle and the rectangle given by the area under straight line graph for constant velocity) cannot be given directly by the product of the time and the average velocity.