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4 years ago

What’s the correct answer

Physics

1 answer:

pshichka [43]4 years ago

6 0

Answer:

The equation for momentum is m x v so multiple 5.9 x 5.0

Explanation:

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During combustion reactions, explain why the energy of the reactants must exceed the total energy of the products

maxonik [38]

Answer:

In these reactions the products are higher in energy than the reactants. ... This barrier is due to the fact that to make CO2 and H2O we have to break 4 carbon-hydrogen bonds and some ...

Explanation:

8 0

3 years ago

Four objects are situated along the y axis as follows: a 2.02 kg object is at 3.03 m, a 2.92 kg object is at 2.43 m, a 2.53 kg o

levacccp [35]

Answer:

0.975 m

Explanation:

The center of mass y = ∑m₁y₁/∑m₁ where m₁ = the individual masses and y₁ = the y - coordinates of the individual masses

So,

y = (2.02 kg × 3.03 m + 2.92 kg × 2.43 m + 2.53 kg × 0 m + 4.02 kg × -0.501 m)/(2.02 kg + 2.92 kg + 2.53 kg + 4.02 kg)

y = (6.1206 kgm + 7.0956 kgm + 0 kgm - 2.01402 kgm)/11.49 kg

y = 11.20218 kgm/11.49 kg

y = 0.975 m

So, the y- component of the center of mass of these objects is 0.975 m

8 0

3 years ago

A block is at rest on the incline shown in the gure. The coefficients of static and kinetic friction are 0.6 and 0.51, respec- t

Alekssandra [29.7K]

Normal reaction force on the block while it is at rest on the inclined plane is given as

$F_n = mgcos\theta$

here we know that

m = 46 kg

$\theta = 29^o$

now we will have

$F_n = 46*9.8*cos29 = 394.3 N$

now the limiting friction or maximum value of static friction on the block will be given as

$F_s = \mu_s * F_n$

$F_s = 0.6 * 394.3 = 236.56 N$

Above value is the maximum value of force at which block will not slide

Now the weight of the block which is parallel to inclined plane is given as

$F_{||} = mg sin\theta$

here we know that

$F_{||} = 46*9.8 sin29 = 218.55 N$

Now since the weight of the block here is less than the value of limiting friction force and also the block is at rest then the frictional force on the block is static friction and it will just counter balance the weight of the block along the inclined plane.

So here <u>friction force on the given block will be same as its component on weight which is 218.55 N</u>

5 0

3 years ago

A 68 kg person runs off a dock at 2.79 m/s and lands on a stationary 131 kg boat. What is their joint velocity afterward?

kolbaska11 [484]

P=mv
pi =pf
(68x2.79)+(131x0) = 199v
v= 0.95 m/s

4 0

3 years ago

Jane plans to fly from Binghamton, New York, to Springfield, Massachusetts, about 280 km due east of Binghamton. She heads due e

aleksklad [387]

Answer:

90m/h N

Explanation:

we consider janes velocity to the positive X axis so Vj=280km/s

and the final position a dot in on the graph r=294km, we find the x and y components:

294*cos(17.8)=280 X+ confirming our theory

294*sin(17.8)=90 Y+ or N and that's the winds' velocity.

5 0

3 years ago