All categories

2 years ago

A car has an initial velocity of 0m/s. It accelerates at a rate of 5m/s2 and travels 30m. What is the final velocity?

Physics

2 answers:

Eduardwww [97]2 years ago

8 0

Answer:

2.5m/s1

Explanation:

Nikolay [14]2 years ago

6 0

Answer:

V=17.3m/ssss

Explanation:

You might be interested in

Six parallel-plate capacitors of identical plate separation have different plate areas a, different capacitances c, and differen

lina2011 [118]

<span>Seis condensadores de placas paralelas de separación de placas idénticas tienen diferentes áreas de placa a, diferentes capacitancias c, y dieléctricos diferentes que llenan el espacio entre las placas. A continuación se muestra un diagrama genérico de lo que cada uno de estos condensadores podría parecer
</span>

6 0

3 years ago

Can someone plz help me

Ymorist [56]

∆ STRONG ACIDS :

• SULPERIC ACID

• HYDROCHLORIC ACID

• NITRIC ACID

∆ STRONG BASES :

• SODIUM HYDROXIDE

• POTASSIUM HYDROXIDE

• CALCIUM HYDROXIDE

• BARIUM HYDROXIDE

∆ WEAK ACIDS :

• ACETIC ACID

• PHOSPERIC ACID

• FORMIC ACID

∆ WEAK BASES :

• ANMONIUM HYDROXIDE

• MAGNESIUM HYDROXIDE

• SODIUM HYDROXIDE

6 0

3 years ago

A basketball player is getting ready to jump, pushing off the ground and accelerating upward.

lidiya [134]

Answer:

Explanation:

3 0

3 years ago

How thick is the Earth's surface?

marissa [1.9K]

It is typically 30 km to 50 km thick.

3 0

4 years ago

Read 2 more answers

Spitting cobras can defend themselves by squeezing muscles around their venom glands to squirt venom at an attacker. Suppose a s

Nezavi [6.7K]

Answer: 1.124 m

Explanation:

This situation is a good example of the projectile motion or parabolic motion, and the main equations that will be helpful in this situations are:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$V_{o}=2.80 m/s$ is the initial speed

$\theta=39\°$ is the angle at which the venom was shot

$t$ is the time since the venom is shot until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=0.4 m$ is the initial height of the venom

$y=0$ is the final height of the venom (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Knowing this, let's begin:

First we have to find $t$ from (2):

$0=0.4 m+2.8 m/s sin(39\°) t+\frac{-9.8m/s^{2}t^{2}}{2}$ (3)

Rearranging (3):

$-4.9 m/s^{2} t^{2} + 1.762 m/s t + 0.4 m=0$ (4)

This is a <u>quadratic equation</u> (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula: