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Sliva [168]
3 years ago
15

Can someone help me with any of these problems or all of them? It will get you an easy ten points per question

Physics
1 answer:
MrRa [10]3 years ago
6 0

Answer: For number 9, the children would not be able to close the box.

Explanation: The combined force of the children is 55 N. The box spring overpowers them with 60 N. I hope this helps!

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A 49 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 3.3 m/s just before hitting the ground.
hichkok12 [17]

Answer:

a) \Delta U_g=-5.3kJ

b) K=0.27kJ

c) F_f=0.45kN

Explanation:

the gravitational potential energy is given by:

U_g=m.g.h\\

\Delta U_g=m.g.h_f-m.g.h_i\\\Delta U_g=49kg*9.8m/s^2*(0m-11m)\\\Delta U_g=-5.3kJ

The kinetic energy is given by:

K=\frac{1}{2}m.v^2\\

the initial kinetic energy is zero because the motion started from rest, so:

K=\frac{1}{2}*49kg*(3.3m/s^2)^2\\K=0.27kJ

applying the conservation of energy theorem:

U_g-W_f=K_f\\W_f=-(\Delta K+\Delta U)\\W_F=5.3kJ-0.27kJ\\W_F=-5.0kJ

The work done by the friction force is given by:

W_f=F_f.h.cos(\theta)\\

the angle of the force is 180 degrees because it's against the movement:

F_f=\frac{W_f}{h.cos(\theta)}\\\\F_f=\frac{-5.0kJ}{11m.cos(180^o)}\\\\F_f=0.45kN

8 0
3 years ago
The product side of a chemical reaction is shown. → 7Ti2(SO4)3
Alex_Xolod [135]
The answer is the fourth choice because there are 7 represents in a coefficient.

7 0
3 years ago
Read 2 more answers
Find the current in the thin straight wire if the magnetic field strength is equal to 0.00005 T at distance 5 cm. ​
Elodia [21]

Answer:

Answer

Correct option is

A

5×10

−6

tesla

I=5A

x=0.2m

Magnetic field at a distance 0.2 m away from the wire.

B=

2πx

μ

0

I

=

2π×0.2

4π×10

−7

×5

=10×5×10

−7

=5×10

−6

tesla

3 0
3 years ago
You have a horizontal grindstone (a disk) that is 86 kg, has a 0.33 m radius, is turning at 92 rpm (in the positive direction),
Rudiy27

Answer:

a) α = 0.338 rad / s²  b)   θ = 21.9 rev

Explanation:

a) To solve this exercise we will use Newton's second law for rotational movement, that is, torque

    τ = I α

    fr r = I α

Now we write the translational Newton equation in the radial direction

    N- F = 0

    N = F

The friction force equation is

    fr = μ N

    fr = μ F

The moment of inertia of a saying is

    I = ½ m r²

Let's replace in the torque equation

    (μ F) r = (½ m r²) α

    α = 2 μ F / (m r)

    α = 2 0.2 24 / (86 0.33)

    α = 0.338 rad / s²

b) let's use the relationship of rotational kinematics

    w² = w₀² - 2 α θ

    0 = w₀² - 2 α θ

    θ = w₀² / 2 α

Let's reduce the angular velocity

     w₀ = 92 rpm (2π rad / 1 rev) (1 min / 60s) = 9.634 rad / s

    θ = 9.634 2 / (2 0.338)

     θ = 137.3 rad

Let's reduce radians to revolutions

    θ = 137.3 rad (1 rev / 2π rad)

    θ = 21.9 rev

7 0
3 years ago
A cup has a mass of 0.0650 kg and a
Roman55 [17]

Answer: 0.185

Explanation:

Trust bro

5 0
2 years ago
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