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inna [77]
3 years ago
7

An OSU linebacker of mass 110.0 kg sacks a UM quarterback of mass 85.0 kg. Just after they collide, they are momentarily stuck t

ogether, and both are moving at a speed of 2.60 m/s. If the quarterback was at rest just before he was sacked, how fast was the linebacker moving just before the collision
Physics
1 answer:
expeople1 [14]3 years ago
3 0

Answer:

The initial speed of the linebacker just before the collision was 4.6 m/s.

Explanation:

Mass of OSU linebacker, m = 110 kg

Mass of UM quarterback, m' = 85 kg

Just after they collide, they are momentarily stuck together, the common speed is, V = 2.6 m/s

Initial speed of quarterback is 0 as it was at rest initially, u' = 0

Let u was the initial speed of the linebacker just before the collision. As they struck together, the momentum remains conserved. So,

mu+m'u'=(m+m')V\\\\mu=(m+m')V\\\\u=\dfrac{(m+m')V}{m}\\\\u=\dfrac{(110+85)\times 2.6}{110}\\\\u=4.6\ m/s

So, the initial speed of the linebacker just before the collision was 4.6 m/s.

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trasher [3.6K]
I believe the answer would be C because potential energy is affected by height and mass. The truck in photo C is the highest and has a lot of mass.
4 0
3 years ago
Suppose that the process were repeated, except that in step 3 a neutral acrylic rod instead of a finger is used to touch the ele
Marta_Voda [28]

Answer:

b) True Only if the finger is isolated from ground

c) True. The total charge does not change since the system is isolated

Explanation:

When the electroscope is touched with an acrylic rod, some charges are transferred from the electroscope to the rod, until the charge in both is equal.

In the case it know when the electroscope is touched with a finger, two things can happen.

- The body is isolated from the ground, the efective charge is redistributed between the two bodies. Case similar to insulating rod

- The body is connected to ground, the charge is transferred to the finger and from here to the ground until the total charge is transferred and the Earth and the final charge of the electroscope is zero.

Let's review the final statements

a) False, when part of the load is touched, it passes to the rod, so when it separates it does not return to the initial load

b) True Only if the finger is isolated from ground

c) True. The total load does not change since the system is isolated

d) False. The value of the load changes =, but its sign does not

8 0
2 years ago
(m/s)<br> 20<br> 10.<br> 100<br> -10<br> total distance
Sergio039 [100]

Answer:

sum of them 4

Explanation:

8 0
3 years ago
To measure the acceleration due to gravity on a distant planet, an astronaut hangs a 0.070-kg ball from the end of a wire. The w
mote1985 [20]

Answer:

g = 1.64m/s²

Explanation:

1.5m in 0.078s

V = 15 / 0.078

= 19.23m/s

Tension = mg

μ = 3.10 × 10⁻⁴

T = V²μ

mg =  V²μ

g =  V²μ / m

g = ((19.23)²(3.10 × 10⁻⁴)) / (0.070)

g = 1.64m/s²

7 0
3 years ago
At a distance of 11 cm from a presumably isotropic, radioactive source, a pair of students measure 65 cps (cps = counts per seco
Alborosie

To solve the problem, it is necessary the concepts related to the definition of area in a sphere, and the proportionality of the counts per second between the two distances.

The area with a certain radius and the number of counts per second is proportional to another with a greater or lesser radius, in other words,

A_1*m=M*A_2

A_i =Area

M,m = Counts per second

Our radios are given by

r_1 = 11cm

R_2 = 20cm

m = 65cps

Therefore replacing we have that,

A_1*m=M*A_2

4\pi r_1^2*m = M * 4\pi R_2^2 M

r^2*m=MR^2

M = \frac{m*r^2}{R^2}

M = \frac{65*11^2}{20^2}

M = 19.6625cps

Therefore the number of counts expect at a distance of 20 cm is 19.66cps

7 0
3 years ago
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