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3 years ago

7

A 150-kg object takes 1.5 minutes to travel a 2,500- meter straight path. It begins the trip traveling 120 m/s and decelerates t

o a velocity of 20 m/s . What is the acceleration?

1 answer:

lana [24]3 years ago

4 0

The answer is it is -1.11 m/s^2.

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In noisy factory environments, it's possible to use a loudspeaker to cancel persistent low-frequency machine noise at the positi

Vaselesa [24]

Answer:

7.39 m or 3.61 m

Explanation:

$\lambda$ = Wavelength

f = Frequency = 90 Hz

v = Speed of sound = 340 m/s

Path difference of the two waves is given by

$s_1-s_2=\frac{\lambda}{2}$

Velocity of wave

$v=f\lambda\\\Rightarrow \lambda=\frac{v}{f}\\\Rightarrow \lambda=\frac{340}{90}\\\Rightarrow \lambda=3.78\ m$

$s_1=s_2\pm\frac{\lambda}{2}\\\Rightarrow s_1=5.5\pm \frac{3.78}{2}\\\Rightarrow s_1=7.39\ m, 3.61\ m$

So, the location from the worker is 7.39 m or 3.61 m

7 0

3 years ago

(I think it's D but idontknow)

sukhopar [10]

During an exothermic reaction; light and heat are released into the environment.

An exothermic reaction is one in which heat is released to the environment. This heat can be physically observed sometimes like in an a combustion reaction.

In an exothermic reaction, the enthalpy of the reactants is greater than the enthalpy of the products.

This heat lost is sometimes felt as the hotness of the vessel in which the reaction has taken place.

In conclusion, light and heat are released into the environment in an exothermic reaction.

Learn more: brainly.com/question/4345448

3 0

2 years ago

Which type of physical activity is being performed in the picture?

mafiozo [28]

B strength training I think that’s the answer

3 0

3 years ago

Read 2 more answers

An Austin volleyball player bumps a 5 kg ball into the air. It reaches a height of 2.8 meters. How fast was the ball going as it

KATRIN_1 [288]

Answer:

v = 7.4 m/s

Explanation:

Given that,

Mass if a volleyball, m = 5 kg

The ball reaches a height of 2.8 m

We need to find how fast the ball is going as it bumped into the air. Ket the velocity is v. Using the conservation of energy to find it as follows :

$mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 2.8} \\\\=7.4\ m/s$

So, the required speed is 7.4 m/s. Hence, the correct option is (b).

6 0

3 years ago

Machmer Hall is 400 m North and 180 m West of Witless.

yan [13]

Answer:

The distance from Witless to Machmer is 438.63 m.

Explanation:

Given that,

Machmer Hall is 400 m North and 180 m West of Witless.

We need to calculate the distance

Using Pythagorean theorem

$D = \sqrt{(d_{m})^2+(d_{w})^2}$

Where, $d_{m}$ =distance of Machmer Hall

$d_{w}$ =distance of Witless

Put the value into the formula

$D = \sqrt{(400)^2+(180)^2}$

$D=438.63\ m$

Hence, The distance from Witless to Machmer is 438.63 m.

5 0

3 years ago