How much force does it take to accelerate a 2000 kg car at 1m/s^2

A spring with constant k = 78 N/m is at the base of a frictionless, 30.0°-inclined plane. A 0.50-kg block is pressed against the

olasank [31]

Explanation:

The given data is as follows.

Spring constant (k) = 78 N/m, $\theta = 30^{o}$

Mass of block (m) = 0.50 kg

According to the formula of energy conservation,

mgh sin $\theta = \frac{1}{2}kx^{2}$

h = $\frac{1}{2} \times \frac{kx^{2}}{mg Sin \theta}$

= $\frac{78 \times 0.04}{2 \times 0.5 \times 9.8 \times 0.5}$

= 0.64 m

Thus, we can conclude that the distance traveled by the block is 0.64 m.

4 0

3 years ago

someone help pls. Two students, Mia and Peter, leave school to meet at the local coffee shop. Peter decides to jog to the coffee

cluponka [151]

Answer:

1) The distance further it takes Peter to arrive at the Coffee shop than Mia is 1.24 km

2) Mia's average speed is 6.00 km/hour

Peter's average speed is 8.48 km/hour

4) Mia's average velocity = Peter's average velocity = 6.00 km/hour

Explanation:

The given information from the diagram are;

The distance Peter jogs from school to the flower shop = 2.00 km

The distance Peter jogs from the Flower shop to the Coffee shop = 2.24 km.

The distance Mia walks from school directly to the Coffee shop = 3.00 km

The time it takes both Peter and Mia to arrive at the coffee shop = 30 minutes = 0.5 hour

1) The total distance Peter travels from school to the Coffee shop = 2.00 km + 2.24 km = 4.24 km

The distance Mia travels from school to the Coffee shop = 3.00 km

The distance further it takes Peter to arrive at the Coffee shop than Mia = 4.24 km - 3.00 km = 1.24 km

The distance further it takes Peter to arrive at the Coffee shop than Mia = 1.24 km

2) $Average \ speed = \dfrac{Total \ distance \ traveled}{Total \ time \ taken \ in \ the \ journey}$

$Therefore, \ Mia's \ average \ speed = \dfrac{3.00 \ km}{0.5 \ hour}= 6.00 \ km/hour$

Mia's average speed = 6.00 km/hour

$Peter's \ average \ speed = \dfrac{4.24 \ km}{0.5 \ hour}= 8.48 \ km/hour$

Peter's average speed = 8.48 km/hour

4) $Average \ velocicty = \dfrac{Displacement }{Time \ taken}$

The displacement from the School to the Coffee shop is 3.00 km for both Mia and Peter

The time it takes both Peter and Mia to arrive at the Coffee shop from the school is 30 minutes = 0.5 hour

$Therefore, \ Mia's \ average \ velocity = \dfrac{3.00 \ km}{0.5 \ hour}= 6.00 \ km/hour$

Mia's average velocity = 6.00 km/hour

$Peter's \ average \ velocity = \dfrac{3.00 \ km}{0.5 \ hour}= 6.00 \ km/hour$

Therefore, Peter's average velocity is also = 6.00 km/hour

6 0

3 years ago

Which of the following is an example of balanced forces?

Sergio [31]

A seesaw remains stationary when two students of equal weight sit on the ends

c

6 0

3 years ago

Read 2 more answers

Unlike the idealized voltmeter, a real voltmeter has a resistance that is not infinitely large. part a a voltmeter with resistan

Margaret [11]

The EMF of the battery includes the force to to drive across its internal resistance. the total resistance:
R = internal resistance r + resistance connected rv
R = r + rv
Now find the current:
V 1= IR
I = R / V1
find the voltage at the battery terminal (which is net of internal resistance) using
V 2= IR
So the voltage at the terminal is:
V = V2 - V1
This is the potential difference vmeter measured by the voltmeter.

6 0

3 years ago

Read 2 more answers

2) Two ice skaters have masses m1 and m2 and are initially stationary. Their skates are identical. They push against one another

worty [1.4K]

Answer:

m_1 / m_2 = sqrt (1 / 2)

Explanation:

Given:

- Initial velocity of both skaters V_i = 0

- Velocity of skater 1 after push = V_1

- Velocity of skater after push = V_2

- Distance traveled by skater 1 = s_1

- Distance traveled by skater 2 = s_2

- s_1 = 2*s_2

- Accelerations of both skaters to halt is equal

Find:

What is the ratio m1/m2 of their masses

Solution:

- Apply conservation of momentum for two skaters just before and after the push as follows:

P_i = P_f

0 = m_1*V_1 - m_2*V_2

- Evaluate: m_1 / m_2 = ( V_2 / V_1 )

- Apply Conservation of Energy on both skaters as follows:

- Skater 1:

0.5*m_1*V_1^2 = u_k*m_1*g*s_1

-Simplify: 0.5*V_1^2 = u_k*g*(2*s_2)

- Skater 2:

0.5*m_2*V_2^2 = u_k*m_2*g*s_2

-Simplify: 0.5*V_2^2 = u_k*g*s_2

- Divide the two energy equations for skaters:

(V_1 / V_2)^2 = 2

(V_2 / V_1)^2 = 1 / 2

- simplify: (V_2 / V_1) = sqrt (1 / 2)

-Hence from earlier momentum conservation results:

m_1 / m_2 = ( V_2 / V_1 ) = sqrt (1 / 2)

6 0

3 years ago