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4 years ago

True or False: when measuring speed the distance must be considered to be rectilinear (in a line)

Physics

2 answers:

baherus [9]4 years ago

4 0

Answer:

I'm pretty sure it's False

Margaret [11]4 years ago

4 0

Answer:

I dont know if I'm correct but I have to go with False

Explanation:

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Two forces act on a 6.00-kg object. One of the forces is 10.0 N. If the object accelerates at 2.00 m/s2

liubo4ka [24]

Given :

Two forces act on a 6.00-kg object. One of the forces is 10.0 N.

Acceleration of object 2 m/s².

To Find :

The greatest possible magnitude of the other force.\

Solution :

Let, other force is f.

So, net force, F = 10 + f.

Now, acceleration is given by :

$a=\dfrac{F}{mass}\\\\a= \dfrac{10+f}{6}\\\\\dfrac{10+f}{6}=2\\\\f = 12 - 10\\\\f = 2 \ N$

Therefore, the greatest possible magnitude of the other force is 2 N.

Hence, this is the required solution.

7 0

3 years ago

4.2 mol of monatomic gas A interacts with 3.2 mol of monatomic gas B. Gas A initially has 9500 J of thermal energy, but in the p

Komok [63]

Answer:

14657.32 J

Explanation:

Given Parameters ;

Number of moles mono atomic gas A , n 1 = 4 .2 mol

Number of moles mono atomic gas B , n 2 = 3.2mol

Initial energy of gas A , K A = 9500 J

Thermal energy given by gas A to gas B , Δ K = 600 J

Gas constant R = 8.314 J / molK

Let K B be the initial energy of gas B.

Let T be the equilibrium temperature of the gas after mixing.

Then we can write the energy of gas A after mixing as

(3/2)n1RT = KA - ΔK

⟹ (3/2) x 4.2 x 8.314 x T = 9500 - 600

T = (8900 x 3 )/(2x4.2x8.314) = 382.32 K

Energy of the gas B after mixing can be written as

(3/2)n2RT = KB + ΔK

⟹ (3/2) x 3.2 x 8.314 x 382.32 = KB + 600

⟹ KB = [(3/2) x 3.2 x 8.314 x 382.32] - 600

⟹ KB = 14657.32 J

6 0

3 years ago

PLEASE SOMEONE ANSWER!!! thank you so so so much!!!!

Oduvanchick [21]

Answer:

As much I know the gravity on moon is 1.62m/s२.

7 0

3 years ago

A uniform ladder 5.0 m long rests against a frictionless, vertical wall with its lower end 3.0 m from the wall. The ladder weigh

DiKsa [7]

Answer:(a)360N,(b)171N,(c)2.702m

Explanation:

(a)Maximum Friction Force = $\mu \left ( N\right )=0.4\times \left ( 740+160\right )$

=360 N

$cos\theta =\frac{3}{5}$

$sin\theta =\frac{4}{5}$

(b)Moment about Ground Point

$740\times 1\times cos\theta +2.5\times 160\times cos\theta -N_15sin\theta$

$N_1tan\theta =1140$

$N_1=171 N$

$N_1=f=171 N$

(c)

$740\times x\times cos\theta +2.5\times 160\times cos\theta -N_15sin\theta$

Here maximum friction force can be 360 N

Therefore $N_1=360 N$

Where x is the maximum distance moved by man along the ladder

$360\times 5\times \frac{4}{3}=740x+160\times 2.5$

740x=2000

x=2.702m

5 0

3 years ago

In general, if the temperature of a chemical reaction is increased, the reaction rateA. IncreasesB. decreasesC. remains the same

In-s [12.5K]

A. Increases

I would assume this to be the answer because heat is another form of energy. If there is more energy the molecules will become more active. This makes A the most logical answer.

6 0

3 years ago