All categories

2 years ago

What is the acceleration of a 1000 kg car subject to a 550 N net force

Physics

1 answer:

Xelga [282]2 years ago

4 0

Answer:

The acceleration of a 1000 kg car subject to a 550 N net force = 0.55 m/s^2

Explanation:

Given:

F = 550 N

m = 1000 kg

To Find:

a = ?

Solution:

So by the equation by Newton's 2nd Law of Motion,

F = m x a

550 N = 1000 kg x a

a = 550 N/ 1000 kg

a = 0.55 m/s^2

Therefore,

The acceleration of a 1000 kg car subject to a 550 N net force = 0.55 m/s^2

PLEASE MARK ME AS BRAINLIEST!!!

You might be interested in

3. The main difference between speed and velocity involves

Zinaida [17]

The difference between them is that
velocity is the speed with a direction, whereas speed does not have a direction.

8 0

3 years ago

What happens when a star exhausts its core hydrogen supply?

Allushta [10]

Answer:

It becomes a giant or supergiant.

Explanation:

Once all the hydrogen supply is gone, fusion of hydrogen into helium stops. The core starts to contract and liberates energy, which heats the superior layer until it becomes hot enough to start the fusion of hydrogen into helium.

6 0

3 years ago

A stationary shell is exploded in to three fragments A, B, C of masses in the ratio 1:2:3. A travels

spin [16.1K]

Answer:

20 m/s

Explanation:

If the mass of fragment A is m, then the mass of fragment B is 2m, and the mass of fragment C is 3m.

The velocity of A is 60 m/s at angle 0°.

The velocity of B is 30 m/s at angle 120°.

The velocity of C is v at angle θ.

In the x direction:

Momentum before = momentum after

(m + 2m + 3m) (0) = m (60 cos 0°) + 2m (30 cos 120°) + 3m (v cos θ)

0 = 60m − 30m + 3m v cos θ

0 = 30m + 3m v cos θ

-30m = 3m v cos θ

-10 = v cos θ

In the y direction:

Momentum before = momentum after

(m + 2m + 3m) (0) = m (60 sin 0°) + 2m (30 sin 120°) + 3m (v sin θ)

0 = 0 + 30√3 m + 3m v sin θ

-30√3m = 3m v sin θ

-10√3 = v sin θ

Square the two equations and add together:

(-10)² + (-10√3)² = (v cos θ)² + (v sin θ)²

100 + 300 = v² cos² θ + v² sin² θ

400 = v² (cos² θ + sin² θ)

400 = v²

v = 20

The speed of fragment C is 20 m/s.

7 0

3 years ago

9) If a wave has a speed of 362 m/s and a period of 4.17 s, what is its wavelength?

kakasveta [241]

Answer:

Option B (1.51 m)

Explanation:

U 2 can help me by marking as brainliest........

5 0

2 years ago

Read 2 more answers

Suppose that you have a 680 Ω, a 720 Ω and a 1.20 kΩ resistor. (a) What is the maximum resistance you can obtain by combining th

Delvig [45]

Explanation:

As the given data is as follows.

$R_{1} = 680 \ohm$ ohm $\ohm$ , $R_{2} = 720 \ohm$ ohm,

$R_{3} = 1.2 k\ohm$ = 1200 $\ohm$ (as 1 k ohm = 1000 m)

(a) We will calculate the maximum resistance by combining the given resistances as follows.

Max. Resistance = $R_{1} + R_{2} + R_{3}$

= $(680 + 720 + 1200) \ohm$ ohm

= 2600 ohm

or, = 2.6 $k\ohm$ ohm

Therefore, the maximum resistance you can obtain by combining these is 2.6 $k\ohm$ ohm.

(b) Now, the minimum resistance is calculated as follows.

Min. Resistance = $\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$

= $\frac{1}{680} + \frac{1}{720} + \frac{1}{1200}$

= $3.683 \times 10^{-3}$ ohm

Hence, we can conclude that minimum resistance you can obtain by combining these is $3.683 \times 10^{-3}$ ohm.

3 0

2 years ago