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3 years ago

A 150 kg car hits a wall with 45 N of force, what was its acceleration?

Physics

1 answer:

alisha [4.7K]3 years ago

4 0

Answer:

a = 0.3 m/s²

Explanation:

Given: 45 N, 150 kg

To find: a

Formula: $a = \frac{F}{m}$

Solution: To find a, divide the force by the weight

A = F ÷ m

= 45 ÷ 150

= 0.3 m/s²

Newtons are derived units, equal to 1 kg-m/s². In other words, a single Newton is equal to the force needed to accelerate one kilogram one meter per second squared.

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Determina la velocidad de un avión que recorre 459 km en 67 min

satela [25.4K]

Answer:

122397.7

Explanation:

4 0

3 years ago

Help!!!, combination circuits, Physics

Kaylis [27]

Current and voltage on each resistor:

$I_1 = 3.98 A, V_1 = 3.98 V$

$I_2=0.015 A, V_2 = 0.075 V$

$I_3 = 0.4 A, V_3 = 0.4 V$

$I_4 = 0.385 A, V_4 = 0.77 V$

$I_5 = 0.585 A, V_5 = 1.17 V$

$I_6 = 3.01 A, V_6 = 6.02 V$

$I_7 = 0.97 A, V_7 = 4.85 V$

Explanation:

In order to solve the circuit, we first have to find the equivalent resistance of the whole circuit, then the total current, and then we can proceed finding the current and the voltage for each resistor.

We start by calculating the equivalent resistance of resistors 2 and 3, which are in parallel:

$R_{23}=\frac{R_2R_3}{R_2+R_3}=\frac{(5)(1)}{5+1}=0.833\Omega$

This resistor is in series with resistor 4, so:

$R_{234}=R_{23}+R_4=0.833+2.0=2.833\Omega$

This resistor is in parallel with resistor 5, therefore:

$R_{2345}=\frac{R_{234}R_5}{R_{234}+R_5}=\frac{(2.833)(2.0)}{2.833+2.0}=1.172\Omega$

This resistor is in series with resistor 7, so:

$R_{23457}=R_{2345}+R_7=1.172+5.0=6.172\Omega$

This resistor is in parallel with resistor 6, so:

$R_{234567}=\frac{R_{23457}R_6}{R_{23457}+R_6}=\frac{(6.172)(2.0)}{6.172+2.0}=1.510\Omega$

Finally, this combination is in series with resistor 1:

$R_{eq}=R_1+R_{234567}=1.0+1.510=2.510\Omega$

We finally found the equivalent resistance of the circuit. Now we can find the total current in the circuit, which is also the current flowing through resistor 1:

$I_1=\frac{V}{R_{eq}}=\frac{10}{2.510}=3.98 A$

And we can also find the potential difference across resistor 1:

$V_1=I_1 R_1=(3.98)(1.0)=3.98 V$

This means that the voltage across resistor 6 is

$V_6=V-V_1=10-3.98=6.02 V$

And so, the current on resistor 6 is

$I_6=\frac{V_6}{R_6}=\frac{6.02}{2.0}=3.01 A$

The current flowing in the whole part of the circuit containing resistors 2,3,4,5,7, and therefore through resistor 7, is

$I_7=I-I_6=3.98-3.01=0.97 A$

And so the voltage across resistor 7 is

$V_7=I_7 R_7=(0.97)(5.0)=4.85 V$

The voltage across resistor 5 is

$V_5 = V_6 - V_7 = 6.02 - 4.85 =1.17 V$

And so the current is

$I_5 = \frac{V_5}{R_5}=\frac{1.17}{2.0}=0.585 A$

The current through resistor 4 is

$I_4 = I_7 - I_5 = 0.97-0.585 = 0.385 A$

And therefore its voltage is

$V_4=I_4 R_4 = (0.385)(2.0)=0.77 V$

So, the voltage through resistor 3 is

$V_3=V_5-V_4=1.17-0.77=0.4 V$

And the current is

$I_3=\frac{V_3}{R_3}=\frac{0.4}{1.0}=0.4 A$

Finally, the current through resistor 2 is

$I_2=I_4-I_3=0.5-0.385=0.015 A$

And so its voltage is

$V_2=I_2R_2=(0.015)(5.0)=0.075 V$

Learn more about current and voltage:

brainly.com/question/4438943

brainly.com/question/10597501

brainly.com/question/12246020

#LearnwithBrainly

4 0

4 years ago

A ball is thrown at 20 m/s from the ground upwards at an angle of elevation of 30°. How far away does it land? 35.35 m

mestny [16]

Answer:

35.35 m

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 20 m/s

Angle of projection (θ) = 30°

Acceleration due to gravity (g) = 9.8 m/s²

Range (R) =.?

The range (i.e how far away) of the ball can be obtained as follow:

R = u² Sine 2θ /g

R = 20² Sine (2×30) / 9.8

R = 400 Sine 60 / 9.8

R = (400 × 0866) / 9.8

R = 346.4 / 9.8

R = 35.35 m

Therefore, the range (i.e how far away) of the ball is 35.35 m

8 0

3 years ago

A baseball, which has a mass of 0.685 kg., is moving with a velocity of 38.0 m/s when it contacts the baseball bat duringwhich t

Evgen [1.6K]

Answers:

a) $65.075 kgm/s$

b) $10.526 s$

c) $61.82 N$

Explanation:

<h3>a) Impulse delivered to the ball</h3>

According to the Impulse-Momentum theorem we have the following:

$I=\Delta p=p_{2}-p_{1}$ (1)

Where:

$I$ is the impulse

$\Delta p$ is the change in momentum

$p_{2}=mV_{2}$ is the final momentum of the ball with mass $m=0.685 kg$ and final velocity (to the right) $V_{2}=57 m/s$

$p_{1}=mV_{1}$ is the initial momentum of the ball with initial velocity (to the left) $V_{1}=-38 m/s$

So:

$I=\Delta p=mV_{2}-mV_{1}$ (2)

$I=\Delta p=m(V_{2}-V_{1})$ (3)

$I=\Delta p=0.685 kg (57 m/s-(-38 m/s))$ (4)

$I=\Delta p=65.075 kg m/s$ (5)

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately $1.0 cm=0.01 m$ :