During a game, a soccer player runs 16.0 m North and then 24.0 m South in 12 s. Find her average speed.

What kind of energy does a light bolt have ?

adell [148]

It has thermal energy

7 0

3 years ago

A Scooter with a mass of 50 kilograms travels with a speed of 10 meters per second. A 2 Nt force bring it to a halt. How long ti

Serga [27]

Answer:

250s

Explanation:

Given parameters:

Mass = 50kg

Speed = 10m/s

Force = 2N

Unknown:

Time the force was acting = ?

Solution:

To solve this problem, we use the equation of motion from the second law of Newton:

Ft = mV

F is the force

t is the time taken

m is the mass

V is the velocity

2 x t = 50 x 10

t = 250s

3 0

3 years ago

Hook's law describes an ideal spring. Many real springs are better described by the restoring force (FSp)s=−kΔs−q(Δs)^3, where q

kvasek [131]

Answer

given,

k = 250 N/m

q = 900 N/m³

(FSp)s=−kΔs−q(Δs)^3

work done = Force x displacement

$W = \int {F. dx}$

limits are x = 0 to x = 0.15 m

work done

$W = \int_0^{0.15} (k x + q x^3)\ dx$

$W = [\dfrac{kx^2}{2}+\dfrac{qx^4}{4}+ C]_0^0.15$

$W = \dfrac{300\times 0.15^2}{2}+\dfrac{900\times 0.15^4}{4}$

W = 3.375 + 0.1139

W = 3.3488 J

b) % cubic term = $\dfrac{0.1139}{3.3488}$

% cubic term = $3.4\ %$

7 0

3 years ago

Difference between matter and energy

Ber [7]

Answer:

Energy is the strength and vitality required for sustained physical or mental activity.

Matter occupies space and possesses rest mass, especially as distinct from energy.

Hope this helps! （づ￣3￣）づ╭❤～

8 0

4 years ago

A 2000-kg truck is being used to lift a 400-kg boulder B that is on a 50-kg pallet A. Knowing the acceleration of the rear-wheel

anzhelika [568]

Answer:

(a) reaction at each front wheel is 5272N (upward)

(b) force between boulder and pallet is 4124N (compression)

Explanation:

Acceleration of the truck $a_{t$ = 1 m/ $s^{2}$ (to the left)

when the truck moves 1 m to the left, the boulder is B and pallet A are raised 0.5 m, then,

$a_{A}$ = 0.5 m/ $s^{2}$ (upward) , $a_{B}$ = 0.5 m/ $s^{2}$ (upward)

Let T be tension in the cable

pallet and boulder: ∑fy = ∑(fy)eff = 2T- ( $m_{A}$ + $m_{B}$ )g = ( $m_{A}$ + $m_{B}$ ) $a_{B}$

= 2T- (400 + 50)*(9.81 m/ $s^{2}$ ) = (400 + 50)*(0.5 m/ $s^{2}$ )

T = 2320N

Truck: $M_{R}$ = ∑( $M_{R}$ )eff: = $-N_{f}$ (3.4m) + $m_{T}$ (2.0m) - T (0.6m)= $m_{T} a_{T}$ (1.0m)

Nf = (2.0m)(2000 kg)(9.81 m/ $s^{2}$ )/3.4m - (0.6 m)(2320 N)/3.4m + (1.0 m)(2000 kg)(1.0 m/ $s^{2}$ ) = 11541.2N - 409.4N - 588.2N = 10544N

∑fy (upward) = ∑(fy)eff: $N_{f}$ + $N_{R}$ - $m_{T}$ g = 0

10544 + $N_{R}$ - (2000kg)(9.81 m/ $s^{2}$ ) = 0

$N_{R}$ = 9076N

∑fx (to the left) = ∑(fx)eff: $F_{R}$ - T = $m_{T} a_{T}$

$F_{R}$ = 2320N + (2000kg)(9.81 m/ $s^{2}$ ) = 4320N

(a) reaction at each front wheel:

1/2 $N_{f}$ (upward): 1/2 (10544N) = 5272N (upward)

(b) force between boulder and pallet:

∑fy (upward) = ∑(fy)eff: $N_{B}$ + $M_{B}$ g - $m_{B}$ $a_{B}$

$N_{B}$ = (400kg)(9.81 m/ $s^{2}$ ) + (400kg)(0.5 m/ $s^{2}$ ) = 4124N (compression)

3 0

4 years ago

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